Section 6.1

499

Chapter 6 Exponential and Logarithmic Functions

6.1 Exponential Functions Section Exercises Verbal

1. Explain why the values of an increasing exponential function will eventually overtake the values of an increasing linear function. Linear functions have a constant rate of change. Exponential functions increase at a rate proportional to their size. So, the bigger they get, the faster they increase, which makes them even bigger, which makes them increase even faster, etc.

2. - 3. The Oxford Dictionary defines the word nominal as a value that is “stated or expressed

but not necessarily corresponding exactly to the real value.”1 Develop a reasonable argument for why the term nominal rate is used to describe the annual percentage rate of an investment account that compounds interest. When interest is compounded, the percentage of interest earned on principal ends up being greater than the annual percentage rate for the investment account, because interest is earned on interest. Thus, the annual percentage rate does not necessarily correspond to the real interest earned, which is the very definition of nominal.

Algebraic For the following exercises, identify whether the statement represents an exponential function. Explain.

4. -

5. A population of bacteria decreases by a factor of 18

every 24 hours.

Exponential functions increase/decrease proportionally, or by a percentage, over the domain, therefore this would be an exponential function; the population decreases by a proportional rate .

6. - 7. For each training session, a personal trainer charges his clients $5 less than the previous

training session. Since this is decreasing at a constant rate, $5, over the domain this would be linear decay, not exponential; the charge decreases by a constant amount each visit, so the statement represents a linear function. .

1OxfordDictionary.http://oxforddictionaries.com/us/definition/american_english/nomina.

Section 6.1

500

8. - For the following exercises, consider this scenario: For each year t, the population of a forest of

trees is represented by the function A(t) = 115(1.025)t . In a neighboring forest, the population of

the same type of tree is represented by the function B(t) = 82(1.029)t . (Round answers to the nearest whole number.)

9. Which forest’s population is growing at a faster rate? The base number that is raised to the exponent is the proportion or percentage increase, therefore the larger base would be the faster growth rate. The larger base is 1.029 compared to 1.025, therefore ( ) 82(1.029)tB t = is growing at the faster rate.

10. - 11. Assuming the population growth models continue to represent the growth of the forests,

which forest will have a greater number of trees after 20 years? By how many?

Evaluating both functions at t = 20 will yield 20(20) 115(1.025) 188.44A = ≈ and 20(20) 82(1.029) 145.25,B = ≈ therefore after 20 years forest A – forest B, or 188 ─ 145 =

43. In 20 years Forest A will have 43 more trees than Forest B. 12. - 13. Discuss the above results from the previous four exercises. Assuming the population

growth models continue to represent the growth of the forests, which forest will have the greater number of trees in the long run? Why? What are some factors that might influence the long-term validity of the exponential growth model? Answers may vary. Initially Forest A has a greater number of trees (115 compared to 82) so for a short period of time forest A will have more trees than Forest B, however in the long run Forest B will surpass growth of Forest A because it is growing at a faster rate. The difference in their rates 1.025 and 1.029 will not start making this change until about 97 years. One way to see this is to graph both functions in your graphing calculator, and find where the point of intersection is. At that point B begins to overtake A in number. Long term validity of the models has several environmental issues that may affect the growth rate , such as what happened in Colorado with a bore infestation or extreme fires destroying several thousands of trees. Answers will vary. Some factors that might influence the long-term validity of the exponential growth model are drought, an epidemic that culls the population, and other environmental and biological factors.

For the following exercises, determine whether the equation represents exponential growth, exponential decay, or neither. Explain.

14. -

Section 6.1

501

15. ( )220 1.06 xy = Exponential functions increase/decrease proportionally, or by a percentage, over the domain, therefore this would be an exponential function. It is exponential GROWTH because the base that is the rate of proportion is greater than 1.

16. -

17. ( )11,701 0.97 ty = Exponential functions increase/decrease proportionally, or by a percentage, over the domain, therefore this would be an exponential function. It is exponential DECAY because the base that is the rate of proportion is .97 AND LESS THAN 1. Also a reminder that the base must be greater than 0; Thus, exponential decay; The decay factor, 0.97, is between 0 and 1.

For the following exercises, find the formula for an exponential function that passes through the two points given.

18. - 19. ( )0, 2000 and (2, 20)

All exponential functions are in the form ( ) ( )xf x a b= , so input 0x = and (0) 2000,f = 0( ) 2000

(1) 20002000,

a baa

=

=

=

Since 2000,a = we solve for b using the second coordinates, when 2,x = (2) 20f = 2

2

2000( ) 20 dividing by 20000.01 take the square root of both sides

0.1

b

bb

=

=

=

Since 2000a = and 0.1b = our equation is ( ) 2000(0.1)xf x = 20. - 21. ( )2,6− and ( )3,1

Because we don’t have the initial value (when x = 0) we substitute both points into an equation of the form ( ) xf x ab= , and then solve the system for a and b . Substituting

( )2,6− gives a first equation of 26 ab−= substituting ( )3,1 gives a second equation of 31 ab= . We first solve for b :

Section 6.1

502

2 2

3 3

2 3

5

5

15

6 6166

16

16

ab a b

ab a b

b b

b

b

b

−

−

−

−

= → =

= → =

=

=

⎛ ⎞ =⎜ ⎟⎝ ⎠

⎛ ⎞ =⎜ ⎟⎝ ⎠

Now use b to solve for :a 3

35

35

1

116

16

ab

a

a−

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠

Finally, substitute your values for a and b into ( ) :xf x ab=

( )35 51 1( ) 2.93 0.699

6 6

x

xf x−

⎛ ⎞ ⎛ ⎞= ≈⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

22. - For the following exercises, determine whether the table could represent a function that is linear, exponential, or neither. If it appears to be exponential, find a function that passes through the points.

23. x 1 2 3 4

f (x) 70 40 10 -20 Since this is decreasing at a constant rate, 30− for each increase of 1 over the domain this would be LINEAR decay, not exponential.

24. - 25.

x 1 2 3 4 m(x) 80 61 42.9 25.61

To find the rate of proportion we may choose any m(x) value and divide it by a preceding

( ).m x If you divide 61/80 =.7625, dividing 42.9/61 = .7032787 or 25.61/42.9 = .59697, therefore this function is not increasing by a percentage rate over the domain. It is not

Section 6.1

503

exponential. It is NOT linear because it is NOT decreasing by a constant rate. Therefore it is NEITHER.

26. - 27.

x 1 2 3 4 g(x) –3.25 2 7.25 12.5

Since this is increasing at a constant rate, 2- (-3.25) = 5.25 for each increase of 1 over the domain this would be linear growth, not exponential

For the following exercises, use the compound interest formula, A(t) = P 1+rn

⎛⎝⎜

⎞⎠⎟

nt

.

• ( )A t is the account value, • t is measured in years, • P is the starting amount of the account, often called the principal, or more

generally present value, • r is the annual percentage rate (APR) expressed as a decimal, and • n is the number of compounding periods in one year.

28. - 29. What was the initial deposit made to the account in the previous exercise?

In the compound interest formula the P represents the initial deposit which is $10,250. 30. - 31. An account is opened with an initial deposit of $6,500 and earns 3.6% interest

compounded semi-annually. What will the account be worth in 20 years? Because our initial deposit is ,P $6,500.P = The interest rate is 3.6 % written as a decimal is 0.036. Since it is compounded semi-annually 2n = and 20 is ,t the number of years it has accumulated. Inputting the correct values into the formula we have

2(20)0.0366,500 12

⎛ ⎞+⎜ ⎟⎝ ⎠

Entering this value into your calculator make sure to use

parentheses. From left to right input the following 6,500 ( 1 + 0.036 ÷ 2) ^ (2*20) , your answer should be $13,268.58 rounded to the nearest cent.

32. - 33. Solve the compound interest formula for the principal, P .

Starting with the formula, since multiplication is occurring between 1ntr

P andn

⎛ ⎞+⎜ ⎟⎝ ⎠

we would be dividing both sides by 1ntr

n⎛ ⎞+⎜ ⎟⎝ ⎠

Section 6.1

504

1( )

1 1

nt

nt nt

rPA t n

r rn n

⎛ ⎞+⎜ ⎟⎝ ⎠=

⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

then rewriting the left hand side with a negative exponent we

write ( ) 1ntr

A t Pn

−⎛ ⎞⋅ + =⎜ ⎟⎝ ⎠

. Therefore ( ) 1ntr

P A tn

−⎛ ⎞= ⋅ +⎜ ⎟⎝ ⎠

34. - 35. How much more would the account in the previous two exercises be worth if it were

earning interest for 5 more years? We found that the initial deposit 5 years ago was $11,002. We will find the compounded amount using this $11,002,P = 12,n = and 0.055r = and 10t = years. When we find that we will take the difference from the amount given in exercise 34 which was

$14,472.74. 12(10)0.055(10) 11,002 1 $19,045.30

12A ⎛ ⎞= + =⎜ ⎟

⎝ ⎠ . Taking the difference

$19,045.30 $14,472.74 $4,572.56− = 36. -

37. Use the formula found in the previous exercise to calculate the interest rate for an account

that was compounded semi-annually, had an initial deposit of $9,000 and was worth $13,373.53 after 10 years.

[1

( ) 1ntA t

r nP

⎡ ⎤⎛ ⎞⎢ ⎥= −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

where 9000,P = (10) 13,373.53,A = semi-annually 2,n = and

10t = years 1

2(10)$13,373.532 1 4%$9,000

r⎡ ⎤⎛ ⎞⎢ ⎥= − =⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥⎣ ⎦

38. - For the following exercises, determine whether the equation represents continuous growth, continuous decay, or neither. Explain.

39. ( )0.753742 ty e= The coefficient to t in the exponent is the growth rate , when that is positive then it is an exponential, or continuous GROWTH function; the growth rate is greater than 0.

40. -

Section 6.1

505

41. ( ) 22.25 ty e −=

The coefficient to t in the exponent is the growth rate , when that is negative then it is an exponential, or continuous DECAY function ; the growth rate is less than 0.

42. - 43. How much less would the account from Exercise 42 be worth after 30 years if it were

compounded monthly instead?

For monthly compounding we use the previous formula A(t) = P 1+rn

⎛⎝⎜

⎞⎠⎟

nt

. with 12. n =

12(30).072(30) 12,000 1 $103,384.2312

A ⎛ ⎞= + =⎜ ⎟⎝ ⎠

. Taking the difference

$104,053.65 $103,384.23 $669.42− = Numeric For the following exercises, evaluate each function. Round answers to four decimal places, if necessary.

44. - 45. 2 3( ) 4 ,xf x += − for ( )1f −

This means to input 1− into the function for x (which is in the exponent). 2( 1) 3( 1) 4 4f − +− = − = −

46. - 47. 1( ) 2 ,xf x e −= − for ( )1f −

This means to input 1− into the function for x (which is in the exponent). ( 1 1)

( 1) 2 .2707f e− −

− = − ≈ − 48. - 49. 2( ) 1.2 0.3,xf x e= − for ( )3f

This means to input 3 into the function for x (which is in the exponent). 2(3)(3) 1.2 0.3 483.8146f e= − ≈

50. - Technology For the following exercises, use a graphing calculator to find the equation of an exponential function given the points on the curve.

51. (0, 3) and (3, 375)

Section 6.1

506

First press [STAT], [EDIT], [1: Edit…], and clear the lists L1 and L2. Next, in the L1 column, enter the x-coordinates, 0 and 3. Do the same in the L2 column for the y-coordinates, 3 and 375 Now press [STAT], [CALC], [0: ExpReg] and press [ENTER]. The display will read

Re* ^

35

Exp gy a b xab

=

=

=

The exponential equation is 3 5xy = ⋅ .

52. - 53. (20, 29.495) and (150, 730.89)

First press [STAT], [EDIT], [1: Edit…], and clear the lists L1 and L2. Next, in the L1 column, enter the x-coordinates, 20 and 150. Do the same in the L2 column for the y-coordinates, 29.495 and 730.89. Now press [STAT], [CALC], [0: ExpReg] and press

[ENTER]. The display is

Re* ^

17.999934331.02500002

Exp gy a b xab

=

=

=

The exponential equation is 18 1.025 xy ≈ ⋅ .

54. -

55. (11, 310.035) and (25, 3563652) First press [STAT], [EDIT], [1: Edit…], and clear the lists L1 and L2. Next, in the L1 column, enter the x-coordinates,11 and 25. Do the same in the L2 column for the y-coordinates, 310.035 and 3563652. Now press [STAT], [CALC], [0: ExpReg] and

press [ENTER]. The display is

Re* ^

.20000000021.950000001

Exp gy a b xab

=

=

=

The exponential equation is 0.2 1.95xy ≈ ⋅ Extensions

56. - 57. Repeat the previous exercise to find the formula for the APY of an account that

compounds daily. Use the results from this and the previous exercise to develop a

Section 6.1

507

function ( )I n for the APY of any account that compounds n times per year. Using the same logic in the previous exercise but compounding it daily would mean that

365.n = It follows that 365365(1)

3651 11 365( ) 365APY 1 1;

365

rr aa aA t a r

a a a

⎡ ⎤⎛ ⎞⎛ ⎞ + −⎢ ⎥⎜ ⎟+ −⎜ ⎟ ⎝ ⎠⎢ ⎥− ⎛ ⎞⎝ ⎠ ⎣ ⎦= = = = + −⎜ ⎟⎝ ⎠

Using the

results from these two examples we would develop the function ( )I n for the APY of any account that compounds n times per year to be:

( ) 1 1nr

I nn

⎛ ⎞= + −⎜ ⎟⎝ ⎠

58. - 59. In an exponential decay function, the base of the exponent is a value between 0 and 1.

Thus, for some number 1,b > the exponential decay function can be written as

1( ) .x

f x ab⎛ ⎞= ⋅ ⎜ ⎟⎝ ⎠

Use this formula, along with the fact that b = en , to show that an

exponential decay function takes the form ( )( ) nxf x a e −= for some positive number n .

Let f be the exponential decay function 1( )x

f x ab⎛ ⎞= ⋅ ⎜ ⎟⎝ ⎠

such that 1.b > Then for some

number 0,n > ( ) ( )( ) ( ) ( )111( ) .

x xx x nxn nf x a a b a e a e a eb

− −− −⎛ ⎞= ⋅ = = = =⎜ ⎟⎝ ⎠

60. - Real-World Applications

61. The fox population in a certain region has an annual growth rate of 9% per year. In the year 2012 , there were 23,900 fox counted in the area. What is the fox population predicted to be in the year 2020? Since the growth rate is a percentage we know that it is growing exponentially. In this exponential function, 23,900 represents the initial fox population, 0.09 represents the growth rate, and 1 0.09 1.09+ = represents the growth factor. We can write this function

as ( )( ) 23,900 1.09 ,xxf x ab= = where x is the number of years that happen. Given the

original year is 2012 and we are asked for the population in year 2020, we enter x as 8

years. Therefore ( )8( ) 23,900 1.09 47,622xf x ab= = = foxes.

62. -

Section 6.1

508

63. In the year 1985, a house was valued at $110,000. By the year 2005, the value had appreciated to $145,000. What was the annual growth rate between 1985 and 2005? Assume that the value continued to grow by the same percentage. What was the value of the house in the year 2010? Since this is an exponential growth problem we use the model ( ) xf x ab= . We start with the given information that initial value $110,000a = and when 20x = years the value has grown to $145,000. Plugging these numbers in we will solve for b

20

20

( )145,000 110,000 110,000

1.31818 201.013908504

xf x ab

b divide by

b take the th root of both sidesb

=

=

=

=

Using 1.013908504b = we can substitute that into the formula and find the value when 25x = years (since it started in 1985 and is going to 2010, that is 25 years)

25(25) 110,000(1.013908504) $155,368.09xf ab= = = Therefore, the annual growth rate ,b written as a percent and rounded to the hundredths

of a percent is 1.39%; and the value in 2010 would be $155,368.09 64. - 65. Jamal wants to save $54,000 for a down payment on a home. How much will he need to

invest in an account with 8.2% APR, compounding daily, in order to reach his goal in 5 years? In exercise #33 we derived the formula for present value from the compound interest

formula ( ) 1ntr

P A tn

−⎛ ⎞= ⋅ +⎜ ⎟⎝ ⎠

using this formula we solve for P when

( ) $54000, 8.2% .082, 365 , 5A t r n daily and t years= = = = =

365(5)0.08254000 1 $35,838.76

365P

−⎛ ⎞= ⋅ + =⎜ ⎟⎝ ⎠

66. - 67. Alyssa opened a retirement account with 7.25% APR in the year 2000. Her initial deposit

was $13,500. How much will the account be worth in 2025 if interest compounds monthly? How much more would she make if interest compounded continuously?

For compounding an investment monthly we the interest formula ( ) 1ntr

A t Pn

⎛ ⎞= +⎜ ⎟⎝ ⎠

where

13,500, 12 monthly, 7.25% .0725 and 25 yearsP n r t= = = = =12(25).0725( ) 13,500 1 $82,247.78

12A t ⎛ ⎞= + =⎜ ⎟

⎝ ⎠

Section 6.1

509

To find the amount compounded CONTINUOUSLY we use the formula ( ) rtA t Pe=

( ) .0725(25)25 13,000 $82,697.53A e= = , Taking the difference of the two amounts

$82,697.53 $82,247.78 $449.75− = . Therefore $82,247.78; $449.75 68. -

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Section 6.2

510

Chapter 6 Exponential and Logarithmic Functions

6.2 Graphs of Exponential Functions Section Exercises Verbal

1. What is an asymptote? What role does the horizontal asymptote of an exponential function play in telling us about the end behavior of the graph? An asymptote is a line that the graph of a function approaches, as x either increases or decreases without bound or approaches from the left or right. The horizontal asymptote of an exponential function tells us the limit of the function’s values as the independent variable gets either extremely large or extremely small.

2. - Algebraic

3. The graph of ( ) 3xf x = is reflected about the y-axis and stretched vertically by a factor of 4. What is the equation of the new function, ( )?g x State its y-intercept, domain, and range. All exponential functions can be written in the form ( ) x cf x ab d+= + . In general a will vertically stretch a graph when 1a > and it will vertically compress the graph when

1a < . Inputting a negative sign in front of the function reflects the graph about the x-

axis and inputting a negative in front of the x variable reflects the graph about the y-axis. The c+ being added to x in the exponent shifts the graph to the left when c is positive and to the right when c is negative. Finally d will shift the graph vertically, up when positive, down when negative. For this graph we are asked to reflect it about the y-axis, thus input a negative in front of

the x variable and stretch it by a factor of 4, thus 4b = and ( )( ) 4 3 ;xg x −= The y-intercept

is found by inputting x = 0, therefore the y-intercept: (0, 4); For domain we consider all x values we may input . There are no restrictions, therefore the Domain is ( ),−∞ +∞ all

real numbers; For range, we consider the values that y involves. The y-values never are below the x-axis therefore the Range is ( )0,+∞ all real numbers greater than 0.

4. -

Section 6.2

511

5. The graph of ( ) 10xf x = is reflected about the x-axis and shifted upward 7 units. What is the equation of the new function, ( )?g x State its y-intercept, domain, and range. For this graph we are asked to reflect it about the x-axis, thus input a negative in front of the function 10x= − , then to shift the graph up 7 units thus 7d = , so our function would be ( ) 10 7;xg x = − + The y-intercept is found by inputting x = 0, therefore the

y-intercept: ( )0, 6 ; For domain we consider all x values we may input . There are no

restrictions, therefore the Domain is ( ),−∞ +∞ all real numbers; For the range, we

consider the values that y outputs. The limit on our y-values before shifting it upward was the x-axis (it is asymptotic to it) therefore the asymptote is at 7y = and the Range is

( ),7−∞ or all real numbers less than 7.

6. -

7. The graph of 21 1( ) 4

2 4

x

f x−

⎛ ⎞= − +⎜ ⎟⎝ ⎠

is shifted left 2 units, stretched vertically by a factor of

4, reflected about the x-axis, and then shifted downward 4 units. What is the equation of the new function, ( )?g x State its y-intercept, domain, and range. For this graph we to shift to the left , which makes 2c+ = + , therefore our exponent will be 2 2x x− + = . Stretching it vertically means that 4a = , multiplying it by 4,we get

1 4 22

− = −g . Reflecting about the x-axis puts a negative in front of the function

making our new 2a = + . Finally shifting the graph down 4 units thus 4d = − , so

inputting that in our model our function would be 1( ) 2 ;4

x

g x ⎛ ⎞= ⎜ ⎟⎝ ⎠

The y-intercept is found

by inputting x = 0, therefore the y-intercept: ( )0 , 2 ; For domain we consider all x values we may input . There are no restrictions, therefore the Domain is all real numbers ( ),−∞ +∞ ; For the range, we consider the values that y outputs. The y-values never are

below the x-axis therefore the Range is ( )0,+∞ all real numbers greater than 0.

Graphical For the following exercises, graph the function and its reflection about the y-axis on the same axes, and give the y-intercept.

8. -

Section 6.2

512

9. ( )( ) 2 0.25 xg x = −

Reflection about the y-axis inputting a negative sign in front of the x, so all positive values of x would be negative and still plotting the y-values the same. the y-intercept remains (0, 2)−

y-intercept: (0, 2)−

10. -

For the following exercises, graph each set of functions on the same axes.

11. 1( ) 3 ,4

x

f x ⎛ ⎞= ⎜ ⎟⎝ ⎠

( )( ) 3 2 ,xg x = and ( )( ) 3 4 xh x =

All exponential functions can be written in the form ( ) x cf x ab d+= + . In general a will

vertically stretch a graph when 1a > and it will vertically compress the graph when

1a < . Inputting a negative sign in front of the function reflects the graph about the x-

axis and inputting a negative in front of the x variable reflects the graph about the y-axis. The c+ being added to x in the exponent shifts the graph to the left when c is positive and to the right when c is negative. Finally d will shift the graph vertically, up when positive, down when negative.

For the graph of 1( ) 3 , 3,4

x

f x a⎛ ⎞= =⎜ ⎟⎝ ⎠

so the y-values will be multiplied by 3 , stretching

it vertically. For the graph of ( )( ) 3 2 , 3,xg x a= = so the y-values will be multiplied by 3 ,

stretching it vertically. For the graph of ( )( ) 3 4 , 3,xh x a= = so the y-values will be multiplied by 3 , stretching it vertically.

Section 6.2

513

12. - For the following exercises, match each function with one of the graphs in

13. ( ) ( )2 0.69 xf x = This would be a choice between A, B or C as the 1base b < means it is decreasing function and comparing with #15 and #18, #15 when x values are negative the y values are lower because of a the larger base and #18 would be higher y-values because of the multiplying factor of 4 having the same base as #13, therefore this has to be choice = B

Section 6.2

514

14. -

15. ( ) ( )2 0.81 xf x = This would be a choice between A, B or C as the 1base b < means it is decreasing function and comparing with #13 and #18, see #13 explanation this choice = A

16. -

17. ( ) ( )2 1.59 xf x = See explanation for # 14 and # 16 which leaves this to be choice = E

18. -

For the following exercises, use the graphs shown. All have the form ( ) .xf x ab=

19. Which graph has the largest value for b? The graph which increases the fastest therefore choice D

20. 21. Which graph has the largest value for a?

The graph with the highest y-intercept, choice C 22. -

For the following exercises, graph the function and its reflection about the x-axis on the same axes.

Section 6.2

515

23. ( )1( ) 42

xf x =

Inputting a negative sign in front of the function reflects the graph about the x-axis. For

every x value you will be plotting the opposite y-value ( )1( ) 42

xg x = −

24. -

25. ( )( ) 4 2 2xf x = − + [Inputting a negative sign in front of the function reflects the graph about the x-axis. For

every x value you will be plotting the opposite y-value ( ) ( )( ) 4 2 2 4 2 2x xg x ⎡ ⎤= − − + = −⎣ ⎦

Section 6.2

516

For the following exercises, graph the transformation of f x( ) = 2x. Give the horizontal

asymptote, the domain, and the range. 26. -

27. ( ) 2 3xh x = + We know that all basic exponential functions are asymptotic to the x-axis unless there is a vertical shift then that horizontal asymptote shifts accordingly. The graph has been shifted up 3, so our horizontal asymptote is now 3y = . There are no restrictions on our x-values, therefore the Domain is ( , )−∞ +∞ or all real numbers and the graph has been shifted up 3 , therefore the Range is (3, )+∞ all real numbers strictly greater than 3.

28. - For the following exercises, describe the end behavior of the graphs of the functions.

29. ( ) ( )5 4 1xf x = − − Since there is a negative in front of our function, this graph is reflected about the x- axis, putting the y-values below the x-axis. It is also shifted down 1 which means the

horizontal asymptote will be at 1y = − . It follows then that as ( ), x f x→∞ →−∞; and

as ( ) , 1 x f x→−∞ →−

30. -

Section 6.2

517

31. ( ) ( )3 4 2xf x −= +

Since the base 1b > this is an increasing function however, this has also reflected about the y-axis which changes it to a decreasing function. It is also shifted up 2 which means

the horizontal asymptote will be at 2y = . It follows then that as ( ) , 2x f x→∞ → ;

and as ( ) , x f x→−∞ →∞

For the following exercises, start with the graph of ( ) 4 .xf x = Then write a function that results from the given transformation.

32. - 33. Shift ( )f x 3 units downward

Shifting downward means 3d = − , therefore ( ) 4 3xf x = −

34. - 35. Shift ( )f x 5 units right

Shifting 5 units to the right means 5c = − , therefore 5( ) 4xf x −= 36. - 37. Reflect ( )f x about the y-axis

Inputting a negative sign in front of the x reflects the graph about the y -axis

( ) 4 xf x −=

For the following exercises, each graph is a transformation of 2 .xy = Write an equation describing the transformation.

38. -

Section 6.2

518

39.

All exponential functions can be written in the form x cy ab d+= + This graph has been reflected about the x-axis, thus a negative needs to be put in front of the function and we see that the horizontal asymptote is shifted up to 3y = , so it has shifted up 3 units making 3d = + Knowing that it has shifted up 3, puts our y-intercept at (0, 1)− prior to that shift up, which means that the multiplying factor 1a = − .

2 3xy = − +

40. -

For the following exercises, find an exponential equation for the graph.

41.

Section 6.2

519

All exponential functions can be written in the form x cy ab d+= + This graph has been reflected about the x-axis, thus a negative needs to be put in front of the function and we see that the horizontal asymptote is shifted up to 7y = , so it has shifted up 7 units making 7d = + Knowing that it has shifted up 7, puts our y-intercept at (0, 2)− prior to that shift up, which means that the multiplying factor 2a = − .

( )2 3 7xy = − +

42. -

Numeric For the following exercises, evaluate the exponential functions for the indicated value of .x

43. ( ) 21( ) 7

3xg x −

= for (6).g

From the graph which is 2( ) 7xf x −= we obtain the y-value when x= 6,

6 2 4(6) 7 7 2401f −= = = , from that we do the vertical compression by multiplying by 13

which is ( ) ( )16 2401 800.33

g = =

44. -

45.

1 1( ) 62 2

x

h x ⎛ ⎞= − +⎜ ⎟⎝ ⎠ for ( 7).h −

From the parent graph which is 1( )2

x

f x ⎛ ⎞= ⎜ ⎟⎝ ⎠

we obtain the y-value when 7x = − ,

71( 7) 1282

f−

⎛ ⎞− = =⎜ ⎟⎝ ⎠

, from that we do the vertical compressing and reflection about the

x-axis by multiplying by 12

− which is ( )1( 7) 128 642

g − = − = − , then we shift it up 6, by

adding 6, 64 6 58− + = − therefore ( 7) 58h − = −

Technology For the following exercises, use a graphing calculator to approximate the solutions of the equation. Round to the nearest thousandth. f (x) = abx + d.

Section 6.2

520

46. -

47.

1 11164 8

x⎛ ⎞= ⎜ ⎟⎝ ⎠

See directions in #47 and Input :

1

2

(1 4)(1 8) ^ ( )116

Y xY

= ÷ ÷

==

Press [WINDOW]. Change y max to include 116. Continue to find the point of

intersection. Your answer should be 2.953x ≈ − 48. -

49. 115 3 2

2

x−⎛ ⎞= −⎜ ⎟⎝ ⎠

See directions in #47 and Input :

1

2

3(1 2) ^ ( 1) 25

Y xY

= ÷ − −

=

You may always hit ZOOM #6 ZSTD, which resets your window to zoom standard 1 10 10 10x to x and y to y= − = = − = . Continue to find the point of intersection.

Your answer should be 0.222x ≈ − 50. -

Extensions

51. Explore and discuss the graphs of ( )( ) xF x b= and 1( ) .

x

G xb⎛ ⎞= ⎜ ⎟⎝ ⎠

Then make a conjecture

about the relationship between the graphs of the functions xb and 1 x

b⎛ ⎞⎜ ⎟⎝ ⎠

for any real

number 0.b >

Because if you were to write 11as b

b− it follows that 11( ) ( )

xx xG x b b

b− −⎛ ⎞= = =⎜ ⎟

⎝ ⎠, this would

then be the reflecting of the graph of ( )( ) xF x b= about the y-axis. thus The graph of

1( )x

G xb⎛ ⎞= ⎜ ⎟⎝ ⎠

is the reflection about the y-axis of the graph of ( ) ;xF x b= For any real

Section 6.2

521

number 0b > and function ( ) ,xf x b= the graph of 1 x

b⎛ ⎞⎜ ⎟⎝ ⎠

is the reflection about the y-

axis, ( ).F x− 52. -

53. Explore and discuss the graphs of ( ) 4 ,xf x = 2( ) 4 ,xg x −= and 1( ) 4 .16

xh x ⎛ ⎞= ⎜ ⎟⎝ ⎠

Then make

a conjecture about the relationship between the graphs of the functions xb and 1 xn b

b⎛ ⎞⎜ ⎟⎝ ⎠

for any real number n and real number 0.b >

Because 2 2 1( ) 4 4 4 416

x x xg x − − ⎛ ⎞= = = ⎜ ⎟⎝ ⎠

g g is the identical function

21 1( ) 4 4 ,

16 4x xh x ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠we see that since the graphs of ( )g x and ( )h x are the same and

are a horizontal shift to the right of the graph of ( );f x For any real number n, real

number 0,b > and function ( ) ,xf x b= the graph of 1 xn b

b⎛ ⎞⎜ ⎟⎝ ⎠

is the horizontal shift

( ).f x n−

54. -

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Section 6.3

522

Chapter 6 Exponential and Logarithmic Functions

6.3 Logarithmic Functions Section Exercises Verbal

1. What is a base b logarithm? Discuss the meaning by interpreting each part of the equivalent equations yb x= and logb x y= for 0, 1.b b> ≠ A logarithm is an exponent. Specifically, it is the exponent to which a base b is raised to produce a given value. In the expressions given, the base b has the same value. The exponent, y, in the expression yb can also be written as the logarithm, log ,b x and the value of x is the result of raising b to the power of .y

2. - 3. How can the logarithmic equation logb x y= be solved for x using the properties of

exponents? Since the equation of a logarithm is equivalent to an exponential equation, the logarithm can be converted to the exponential equation ,yb x= and then properties of exponents can be applied to solve for x.

4. - 5. Discuss the meaning of the natural logarithm. What is its relationship to a logarithm with

base b, and how does the notation differ? The natural logarithm is a special case of the logarithm with base b in that the natural log always has base .e Rather than notating the natural logarithm as ( )log ,e x the notation

used is ( )ln .x

Algebraic For the following exercises, rewrite each equation in exponential form.

6. - 7. log ( )a b c=

Rewriting logbase N exponent= in exponential form is exponentbase N= , since the base =

a and the exponent is c , we rewrite this ca b= 8. - 9. ( )log 64x y=

Rewriting logbase N exponent= in exponential form is exponentbase N= , since the base =

x and the exponent is y , we rewrite this 64yx = 10. -

Section 6.3

523

11. ( )15log a b=

Rewriting logbase N exponent= in exponential form is exponentbase N= , since the base = 15 and the exponent is b , we rewrite this 15b a=

12. - 13. ( )13log 142 a=

Rewriting logbase N exponent= in exponential form is exponentbase N= , since the base = 13 and the exponent is a , we rewrite this 13 142a =

14. - 15. ( )ln w n=

Rewriting logbase N exponent= in exponential form is exponentbase N= , but since there

is ( )ln w n= this means the same thing as loge w n= natural logarithm base which = e

and the exponent is n , we rewrite this ne w= For the following exercises, rewrite each equation in logarithmic form.

16. - 17. dc k=

Rewriting an exponential expression exponentbase N= in logarithmic form is logbase N exponent= . Since the base c= and the exponent (which always is alone in the logarithmic equation), exponent d= , we rewrite this = log ( )c k d=

18. - 19. 19x y=

Rewriting an exponential expression exponentbase N= in logarithmic form is logbase N exponent= . Since the 19base = and the exponent (which always is alone in the logarithmic equation), exponent x= , we rewrite this 19log y x=

20. - 21. 4 103n =

Rewriting an exponential expression exponentbase N= in logarithmic form is logbase N exponent= . Since the base n= and the exponent (which always is alone in the

logarithmic equation), 4exponent = , we rewrite this ( )log 103 4n =

22. -

23. 39100

xy =

Rewriting an exponential expression exponentbase N= in logarithmic form is

Section 6.3

524

logbase N exponent= . Since the base y= and the exponent (which always is alone in the

logarithmic equation), exponent x= , we rewrite this 39log100y x⎛ ⎞ =⎜ ⎟⎝ ⎠

24. 25. ke h=

Rewriting an exponential expression exponentbase N= in logarithmic form is logbase N exponent= . Since the base e= , we use ln which means loge . The exponent (which always is alone in the logarithmic equation), exponent k= , we rewrite this

( )ln h k=

For the following exercises, solve for x by converting the logarithmic equation to exponential form.

26. - 27. 2log ( ) 3x = −

Rewriting logbase N exponent= in exponential form is exponentbase N= , since the base =

2 and the exponent is 3− , we rewrite this 32 x− = , therefore 3 128

x −= =

28. - 29. ( )3log 3x =

Rewriting logbase N exponent= in exponential form is exponentbase N= , since the base = 3 and the exponent is 3 , we rewrite this 33 x= , therefore 33 27x = =

30. -

31. 91log ( )2

x =

Rewriting logbase N exponent= in exponential form is exponentbase N= , since the base =

9 and the exponent is 12

, we rewrite this 129 x= , therefore

12 129 9 3x = = =

32. - 33. ( )6log 3x = −

Rewriting logbase N exponent= in exponential form is exponentbase N= , since the base =

6 and the exponent is 3− , we rewrite this 36 x− = , therefore 33

1 162166

x −= = =

34. -

Section 6.3

525

35. ( )ln 2x =

Rewriting logbase N exponent= in exponential form is exponentbase N= , since there is an ln written ,the base = e and the exponent is 2 , we rewrite this 2e x= , therefore 2x e= . (Note: we leave this with e in our answer as it is the exact answer. To approximate it you may use the calculator and obtain 2 7.389056099e ≈ )

For the following exercises, use the definition of common and natural logarithms to simplify.

36. - 37. ( )log 3210

Let log(32) ,x= that means 10 32x = , so ( )log 3210 10 32x= = 38. - 39. ( )ln 1.06e

Let ln(1.06) ,x= rewriting it exponentially yields 1.06xe = , so ( )ln 1.06 1.06xe e= = 40. - 41. ( )ln 10.125 4e +

Let ln(10.125) ,x= rewriting it exponentially yields 10.125xe = , so ( )ln 10.125 10.125xe e= = , therefore ( )ln 10.125 4 10.125 4 14.125e + = + =

Numeric For the following exercises, evaluate the base b logarithmic expression without using a calculator.

42. - 43. ( )6log 6

Let ( )6log 6 x= , then rewrite it in exponential form , since the base = 6 and the

exponent is x , we rewrite this 1

2 1 26 6 6x = = . Therefore x = 12

; ( )61log 62

=

44. - 45. ( )86log 4

Let ( )8log 4 x= , then rewrite it in exponential form , since the base = 8 and the

exponent is x , we rewrite this 8 4x = rewriting each side with the a base of 2 we have

3 2

3 2

8 4(2 ) 22 2

x

x

x

=

=

=

Section 6.3

526

It follows that 23 23

x so x= = , Since ( )82log 43

x= = , evaluating

6ilog8 4( ) = 6i 23⎛

⎝⎜⎞

⎠⎟= 4

For the following exercises, evaluate the common logarithmic expression without using a calculator.

46. - 47. ( )log 0.001

Let log(.001) ,x= that means 33

1 110 .001 101,000 10

x −= = = = , that means 3x = − , so

log(.0001) 3= − . 48. - 49. ( )32log 100−

Let ( )3log 100 x− = , then rewrite it in exponential form yb x= , since the base = 10 and

the exponent is x , we rewrite this ( ) 33 2 610 100 10 10x −− −= = = . It follows that 6,x = −

therefore ( )3log 10 6− = − . Evaluating ( )32log 100 2( 6) 12− = − = −

For the following exercises, evaluate the natural logarithmic expression without using a calculator.

50. - 51. ( )ln 1

Let ln(1) ,x= rewriting it exponentially yields 01xe e= = , since the bases are the same the exponents must be equal, therefore x must equal 0; ln(1) 0=

52. -

53. 2525ln e

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

Let 25ln( ) ,e x= rewriting it exponentially yields

25xe e= , since the bases are the same the

exponents must be equal, therefore 25

x = ; 25 225ln 25 10

5e⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

Technology

Section 6.3

527

For the following exercises, evaluate each expression using a calculator. Round to the nearest thousandth.

54. - 55. ( )ln 15

( )ln 15 2.708050201= rounded to the thousandths place 2.708≈

56. - 57. ( )log 2

log( (2)) .1505149978= rounded to the thousandths place 0.151≈ 58. -

Extensions

59. Is 0x = in the domain of the function ( ) log( )?f x x= If so, what is the value of the function when 0?x = Verify the result. To verify we the end conclusion stated here, we input zero in for x and rewriting

log(0)N = in exponential form we obtain 10 0,N = Since there are no possible real solutions when raising 10 to the N power that can result in an answer of zero, zero must be eliminated from the domain. No, the function has no defined value for 0.x = Therefore, 0x = is not the domain of the function ( ) log( ).f x x=

60. - 61. Is there a number x such that ln 2?x = If so, what is that number? Verify the result.

Yes. Suppose there is a real number, x , such that ln 2.x = Rewriting as an exponential equation gives 2 ,x e= which is a real number ( 2 7.389056099x e= ≈ ). To verify, let 2.x e=

( )ln ln7.389056099 2.x = = Then, by definition, ( ) ( )2ln ln 2.x e= =

62. -

63. Is the following true: ( )( )

1.725ln1.725?

ln 1

e= Verify the result.

Since ( )ln 1 0,= we would have division by zero and this would be an undefined expression.

Therefore , no it is not true; The expression ( )( )

1.725ln

ln 1

e is undefined.

Real-World Applications

64. -

65. Refer to the previous exercise. Suppose the light meter on a camera indicates an EI of 2,− and the desired exposure time is 16 seconds. What should the f-stop setting be?

Section 6.3

528

Inserting the given numbers in their respective places in the formula, we will continue to

solve for f ; 2

22 log16f⎛ ⎞

− = ⎜ ⎟⎝ ⎠

. Rewriting this exponentially yields

22

2

2

2

216

1 multiply by164 16

116 164 16

42

f

f

f

ff

− =

=

=

=

± =

g g

We choose the positive value, therefore 2f = 66. -

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Section 6.4

529

Chapter 6 Exponential and Logarithmic Functions

6.4 Graphs of Logarithmic Functions Section Exercises Verbal

1. The inverse of every logarithmic function is an exponential function and vice-versa. What does this tell us about the relationship between the coordinates of the points on the graphs of each? Since the functions are inverses, their graphs are mirror images about the line .y x= So for every point ( , )a b on the graph of a logarithmic function, there is a corresponding point ( , )b a on the graph of its inverse exponential function.

2. - 3. What type(s) of translation(s), if any, affect the domain of a logarithmic function?

Shifting the function right or left and reflecting the function about the y-axis will affect its domain.

4. - 5. Does the graph of a general logarithmic function have a horizontal asymptote? Explain.

No. A horizontal asymptote would suggest a limit on the range, and the range of any logarithmic function in general form is all real numbers.

Algebraic For the following exercises, state the domain and range of the function.

6. -

7. 1( ) ln2

h x x⎛ ⎞= −⎜ ⎟⎝ ⎠

We must keep

1 021 0212

x

x x x

x

− >

− + > +

>

it follows that 12

x < , therefore the Domain is −∞, 12

⎛⎝⎜

⎞⎠⎟

; The y-values then take on all

values with no restriction, therefore the Range is −∞, ∞( )

8. -

Section 6.4

530

9. ( )( ) ln 4 17 5h x x= + −

We must keep

4 17 04 17 17 0 17

4 174 174 4

xx

xx

+ >

+ − > −

> −

−>

it follows that 174

x−

> , therefore the Domain is −174

, ∞⎛⎝⎜

⎞⎠⎟

; The y-values then take on

all values with no restriction, therefore the Range is −∞, ∞( )

10. -

For the following exercises, state the domain and the vertical asymptote of the function.

11. ( ) ( )log 5f x x= −

For the domain, we must keep

5 0

5 5 0 55

xx

x

− >

− + > +

>

it follows that 5x > , therefore the Domain is 5, ∞( ); This graph shifts to the right 5

units, therefore the vertical asymptote which is normally occurs at 0x = , shifts to the right 5 units. The vertical asymptote will be at 5x = .

12. -

13. ( ) ( )log 3 1f x x= +

For the domain, we must keep

3 1 03 1 1 0 1

3 13 13 3

13

xx

xx

x

+ >

+ − > −

> −

−>

−>

Section 6.4

531

it follows that 13

x−

> , therefore the Domain is − 13

, ∞⎛⎝⎜

⎞⎠⎟

; This graph shifts to the left 13

units, therefore the vertical asymptote which is normally occurs at 0x = , shifts to the left 13

units. The vertical asymptote will be at 13

x = − .

14. - 15. ( ) ( )ln 3 9 7g x x= − + −

For the domain, we must keep

3 9 03 9 9 0 9

3 93 93 3

3

xx

xx

x

+ >

+ − > −

> −

−>

> −

it follows that 3x > − , therefore the Domain is −3, ∞( ); This graph reflects about the x-

axis and shifts to the left 3 units, therefore the vertical asymptote which is normally occurs at 0x = , shifts to the left 3 units. The vertical asymptote will be at 3x = − .

For the following exercises, state the domain, vertical asymptote, and end behavior of the function.

16. -

17. 3( ) log7

f x x⎛ ⎞= −⎜ ⎟⎝ ⎠

For the domain, we must keep

3 07

3 3 307 7 7

37

x

x

x

− >

− + > +

>

it follows that 37

x > , therefore the domain is 37

, ∞⎛⎝⎜

⎞⎠⎟

; This graph shifts to the right 37

units, therefore the vertical asymptote which is normally occurs at 0x = , shifts to the

right 37

units. The vertical asymptote will be at 37

x = . Sketching this graph helps

describe the end behavior.

Section 6.4

532

End behavior: as x approaches 37

from the right, determine what y-value, ( ),f x

approaches. 3 , ( )7

x f x+

⎛ ⎞→ → −∞⎜ ⎟⎝ ⎠

and as x approaches +∞ , determine what y-value,

( ),f x approaches. As , ( )x f x→∞ →∞

18. - 19. ( )( ) ln 2 6 5g x x= + −

For the domain, we must keep

2 6 02 6 6 0 6

2 62 62 2

3

xx

xx

x

+ >

+ − > −

> −

−>

> −

it follows that 3x > − , therefore the domain is −3, ∞( ); This graph shifts to the left 3

units, therefore the vertical asymptote which is normally occurs at 0x = , shifts to the right 3− units. The vertical asymptote will be at 3x = − . Sketching this graph helps describe the end behavior. End behavior: as x approaches 3− from the right, determine what y-value, ( ),f x approaches. 3 , ( )x f x+→− →−∞ and as x approaches +∞ , determine what y-value,

( ),f x approaches. As , ( )x f x→∞ →∞

20. - For the following exercises, state the domain, range, and x- and y-intercepts, if they exist. If they do not exist, write DNE.

21. ( )4( ) log 1 1h x x= − +

For the domain, we must keep

1 0

1 1 0 11

xx

x

− >

− + > +

>

it follows that 1x > , therefore the domain is 1, ∞( ); The y-values then take on all values

with no restriction, therefore the range is −∞, ∞( ) . This graph shifts to the right 1 unit,

therefore the vertical asymptote which is normally occurs at 0x = , will be at 1x = . To find the x-intercept we set y = 0:

Section 6.4

533

( )( )( )

4

4

41

0 log 1 10 1 log 1 1 1

1 log 14 1

1 1 1 14

54

xxx rewrite exponentially

x

x

x

−

= − +

− = − + −

− = −

= −

+ = − +

=

Therefore the x-intercept is 5 , 0 ;4

⎛ ⎞⎜ ⎟⎝ ⎠

To find the y-intercept we set x = 0:

( )( )

4

4

log 0 1log 1

4 1y

yy no real solution

or rewrite exponentially

no real solution

= −

= −

= −

Therefore the y-intercept DNE.

22. - 23. ( )( ) ln 2g x x= − −

For the domain, we must keep

00

1 10

xx

x

− >

−<

− −<

it follows that 0x < , therefore the domain is −∞, 0( ); The y-values then take on all

values with no restriction, therefore the range is −∞, ∞( ) . This graph is not shifted,

therefore the vertical asymptote is at 0x = . To find the x-intercept we set y = 0:

( )( )( )

2

2

2

0 log 20 2 log 2 2

2 log rewrite exponentially

1 1

e

e

e

xxx

e xe x

e x

= − −

+ = − − +

= −

= −−

=− −− =

Therefore the x-intercept is ( )2 , 0 ;e−

To find the y-intercept we set x = 0:

Section 6.4

534

( )( )

log 0 2log 0 no real solution

or rewrite exponentially0

no real solution

e

e

y

yy

e

= −

=

=

Therefore the y-intercept DNE. 24. - 25. ( )( ) 3ln 9h x x= −

For the domain, we must keep 0x > it follows that 0x > , therefore the Domain is 0, ∞( ); The y-values then take on all values

with no restriction, therefore the Range is −∞, ∞( ) . This graph does not shift, therefore

the vertical asymptote remains at 0x = . To find the x-intercept we set y = 0:

( )( )( )( )

( )3

0 3log 90 9 3log 9 9

9 3log3log9

3 33 log

e

e

e

e

e

xxxx

x rewrite exponentiallye x

= −

+ = − +

=

=

=

=

Therefore the x-intercept is ( )3 , 0 ;e

To find the y-intercept we set x = 0:

( )( )

3log 0 9log 0 no real solution

or rewrite exponentially0

no real solution

e

e

y

yy

e

= −

=

=

Therefore they-intercept DNE. Graphical For the following exercises, match each function with the letter corresponding to its graph.

Section 6.4

535

26. - 27. ( ) ln( )f x x=

Since this base is e and 2.71828e ≈ it would be the second lowest base to # 28, therefore it is choice = B

28. - 29. ( )5( ) logh x x=

Since this base is 5 it would be the third lowest base to bases 2 and e, therefore it is choice = C

30. - For the following exercises, match each function with the letter corresponding to its graph.

Section 6.4

536

31. ( )13

( ) logf x x=

Since the base 1 13

< , this graph will be reflected about the x-axis and since 1 33 4

<

comparing it to the graph of #33, the smaller base will decrease slower, therefore this graph is choice = B

32. - 33. ( )3

4

( ) logh x x=

Since the base 3 14

< , this graph will be reflected about the x-axis and since 1 33 4

<

comparing it to the graph of #31, this being the larger base will decrease faster, therefore this graph is choice = C

For the following exercises, sketch the graphs of each pair of functions on the same axis.

34. - 35. ( ) log( )f x x= and 1

2

( ) log ( )g x x=

From the first graph we have base 10 1> , which would have the normal logarithm model.

The second graph with a base 1 12

< will be reflected about the x-axis and decreasing at a

faster rate than the reflected graph of the first which would be a base of 110

.

Section 6.4

537

36. -

37. ( ) xf x e= and ( ) ln( )g x x=

These graphs are simply inverses of one another. Plot the exponential function ( ) xf x e=and interchange the x and y values to plot the second graph. For example ( )( )( )1 1

0,1 (1,0)

1, ( ,1)

1, ( , 1)e e

would be

e would be e

would be− −

Also these would be mirror images of each other reflected about the x = y axis.

For the following exercises, match each function with the letter corresponding to its graph.

38. -

Section 6.4

538

39. ( )4( ) log 2g x x= − +

This should be the graph of ( )4( ) logf x x= reflected about the x-axis axis making the x-

intercept remain as ( )1,0 and shifted left 2 units, making the x-intercept shift from

( ) ( )1,0 1,0to − . Therefore choice = C

40. - For the following exercises, sketch the graph of the indicated function.

41. ( ) 2log ( 2)f x x= +

This should be the graph of ( )2( ) logf x x= shifted left 2 units, making the x-intercept shift

from ( ) ( )1,0 1,0to − .

42. -

43. ( ) ( )lnf x x= −

This should be the graph of ( )( ) lnf x x= reflected about the y-axis, making the x-intercept

shift from ( ) ( )1,0 1,0to − .

Section 6.4

539

44. - 45. ( )( ) log 6 3 1g x x= − +

For the domain, we must keep

6 3 06 3 6 0 6

3 63 63 3

2

xx

xx

x

− >

− − > −

− > −

− −<

− −<

it follows that since the domain is ( ) , 2 ;−∞ This graph shifts to the right 2 units, therefore the vertical asymptote which normally occurs at 0x = , shifts to the right 2 units and is at 2x = . It also has been shifted up 1 unit. To find the x-intercept we set y = 0:

( )( )( )

1

1

0 log 6 3 10 1 log 6 3 1 1

1 log 6 310 6 3

10 6 6 3 65.9 35.9 33 3

1.96661.96

xxx rewrite exponentially

xx

xx

xx

−

−

= − +

+ − = − + −

− = −

= −− = − −− = −− −

=− −

=≈

Therefore the x-intercept is ( )1.96, 0 ;

To find the y-intercept we set x = 0:

Section 6.4

540

( )( )

log 6 3(0) 1log 6 1.7781512504 11.8

yyyy

= − +

= += +≈

Therefore the y-intercept is ( )0, 1.8

46. - For the following exercises, write a logarithmic equation corresponding to the graph shown.

47. Use 2log ( )y x= as the parent function.

Using the model given in the text: ( )logby a x c d= + +

This graph has been reflected about the y-axis thus we put a negative sign in front of the

x in the function. We also see that the x-intercept has shifted from ( ) ( )1,0 0,0to−which means shifting 1 unit to the right would make c+ in our model be 1c+ = − , therefore our equation would be ( )( )2log 1y x= − −

Section 6.4

541

48. -

49. Use 4( ) log ( )f x x= as the parent function.

Using the model given in the text: ( )logby a x c d= + +

This graph has NOT been reflected about either axes. We also see that the x-intercept has shifted from ( ) ( )1,0 1,0to − which means shifting 2 units to the left would make

c+ in our model be 2c+ = + . This graph has a vertical asymptote at –2x = . We know it has no vertical shift because it stayed on the x-axis therefore there may be a vertical

stretch. Now to find a we substitute any coordinates we see on the graph such as ( )0,1.5into our model:

( )( )

( ) ( )

( )

4

4 4

log1.5 log 0 2

11.5 log 2 log 22

1.5 .51.5 (.5).5 .53

by a x ca

a is

aa

a

= +

= +

=

=

=

=

Therefore our equation would be ( )43log 2y x= + .

50. -

Section 6.4

542

Technology For the following exercises, use a graphing calculator to find approximation solutions to each equation.

51. ( ) ( )log 1 2 ln 1 2x x− + = − +

Enter each function into your graphing calculator ( )

( )log 1 2

ln 1 2

x

x

= − +

= − +

1

2

Y

Y

2nd calculate “intersection” and round answer to the thousandths place 2x = 52. - 53. ( ) ( )ln 2 ln 1x x− = − +

( )( )

ln 2

ln 1

x

x

= −

= − +

1

2

Y

Y

2nd calculate “intersection” and round answer to the thousandths place 2.303x ≈ 54. -

55. ( ) ( )1 1log 1 log 13 3

x x− = + +

( )

( )

1 log 13

1log 13

x

x

= −

= + +

1

2

Y

Y

2nd calculate “intersection” and round answer to the thousandths place x ≈ −0.472 Extensions

56. -

57. Explore and discuss the graphs of ( )12

( ) logf x x= and ( )2( ) log .g x x= − Make a

conjecture based on the result. The graphs of ( )1

2

( ) logf x x= and ( )2( ) logg x x= − appear to be the same; Conjecture: for

any positive base 1,b ≠ ( ) ( )1log log .bb

x x= −

58. -

59. What is the domain of the function 2( ) ln ?4

xf x

x+⎛ ⎞= ⎜ ⎟−⎝ ⎠

Discuss the result.

Section 6.4

543

Recall that the argument of a logarithmic function must be positive, so we determine

where 2 04

xx

+>

−. For the domain 2 0

4xx

+>

−, this is true when both 2 0x + > and 4 0x − > ,

which would be intersection of the two sets 2 4,x and x> − > which would be 4x > . Of course it would also be true when both 2 0x + < and 4 0x − < , which would be intersection of the two sets 2 4,x and x< − < which would be 2x < − . So the domain is

restricted to 2 4x or x< − > . Domain −∞, −2( )∪ 4, ∞( ).

From the graph of the function f x( ) =x + 2x − 4

, note that the graph lies above the x-axis on

the interval −∞, −2( ) and again to the right of the vertical asymptote, that is 4, ∞( ).

Therefore, the domain is −∞, −2( )∪ 4, ∞( ).

60. -

Thisfileiscopyright2015,RiceUniversity.AllRightsReserved.

Section6.5

544

Chapter 6 Exponential and Logarithmic Functions

6.5 Logarithmic Properties Section Exercises Verbal

1. How does the power rule for logarithms help when solving logarithms with the form

( )log ?nb x

Any root expression can be rewritten as an expression with a rational exponent so that the power rule can be applied, making the logarithm easier to calculate. Thus, ( )log n

b x can

be written as ( )1 1log log .n

b bx xn

⎛ ⎞=⎜ ⎟⎜ ⎟

⎝ ⎠

2. - Algebraic For the following exercises, expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.

3. ( )log 7 2b x y⋅

( )log 7 2 log (7 ) log (2 )b b bx y x y⋅ = + , then each term is the product rule again, where

( )log 7 log (7) log ( )b b bx x⋅ = + and ( )log 2 log (2) log ( )b b by y⋅ = + , So the final expansion is

( ) ( ) ( ) ( )log 2 log 7 log logb b b bx y+ + +

4. -

5. 13log17b⎛ ⎞⎜ ⎟⎝ ⎠

The quotient rule for logs is subtraction for individual logs therefore 13log17b⎛ ⎞⎜ ⎟⎝ ⎠

=

( ) ( )log 13 log 17b b−

6. -

7. 1ln4k

⎛ ⎞⎜ ⎟⎝ ⎠

1ln ln1 ln(4 ) 0 ln(4 ) ln(4 )4

k k kk

⎛ ⎞ = − = − = −⎜ ⎟⎝ ⎠

then using the power rule the final expansion is

ln(4)k− 8. -

Section6.5

545

For the following exercises, condense to a single logarithm if possible.

9. ( ) ( ) ( )ln 7 ln lnx y+ + Since addition is going on as individual logarithms, using the product rule in reverse, condensing to a single logarithm will result in multiplication, therefore

( ) ( ) ( )ln 7 ln lnx y+ + would be equal to ( )ln 7 x yg g = ( )ln 7xy

10. - 11. ( ) ( )log 28 log 7b b−

Since subtraction is going on as individual logarithms, using the quotient rule in reverse, condensing to a single logarithm will result in division, therefore

( ) ( ) 28log 28 log 7 log7b b b

⎛ ⎞− = ⎜ ⎟⎝ ⎠

= ( )log 4b

12. -

13. 1log7b⎛ ⎞− ⎜ ⎟⎝ ⎠

Rewriting this as individual logs because of division would be subtraction , thus

( ) ( )( ) ( )( )1log log 1 log 7 0 log 77b b b b⎛ ⎞− = − − = − −⎜ ⎟⎝ ⎠

= ( )log 7b .

14. -

For the following exercises, use the properties of logarithms to expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.

15. 15 13

19log x yz

⎛ ⎞⎜ ⎟⎝ ⎠

Applying the product rule to the numerator will result in addition and applying the quotient rule to the denominator will result in subtraction we have

( ) ( ) ( )15 13

15 13 1919log log log logx y

x y zz

⎛ ⎞= + −⎜ ⎟

⎝ ⎠ next we apply the power rule and each of the

exponents become coefficients, therefore the final expansion will be( ) ( ) ( )15log 13log 19logx y z+ −

16. -

17. ( )3 4log x y−

Rewriting this and then applying the product rule we have

( ) ( ) ( )1 3 3

3 4 3 4 2 22 2 2log log log log logx y x y x y x y− − − −⎛ ⎞ ⎛ ⎞= = = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

g g . Next we apply the power

Section6.5

546

rule and each of the exponents become coefficients, therefore the final expansion will be

= ( )3 log 2log( )2

x y− .

18. -

19. ( )2 3 2 53log x y x y

Rewriting this, and simplifying we have

( ) ( )1 5 52 2

2 3 2 5 2 3 2 5 2 3 2 333 3 3 3 3

5 82 142 33 3 3 3

log log log log

log log

x y x y x y x y x y x y x x y y

x y x y+ +

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = ⋅ ⋅ ⋅⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= ⋅ = ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

g g

Now we will apply the product rule and power rules 8 143 3log x y⎛ ⎞⋅⎜ ⎟

⎝ ⎠ = ( ) ( )8 14log log

3 3x y+ .

For the following exercises, condense each expression to a single logarithm using the properties of logarithms.

20. - 21. ( ) ( )9 2ln 6 ln 3x x−

Since subtraction is going on as individual logarithms, using the quotient rule in reverse, condensing to a single logarithm will result in division, therefore

( ) ( ) ( )9

9 2 9 22

6ln 6 ln 3 ln ln 23x

x x xx

−⎛ ⎞− = =⎜ ⎟

⎝ ⎠ = ( )7ln 2x

22. -

23. ( ) ( ) ( )1log log 3log2

x y z− +

Since there are coefficients on the front of the logarithms we apply the power rule first

and place them as exponents ( ) ( ) ( ) ( ) ( )1

321log log 3log log log log2

x y z x y z⎛ ⎞− + = − +⎜ ⎟⎝ ⎠

Next

we group the first two terms and apply the quotient rule, then product rule and condensing to a single logarithm will result in

( ) ( ) ( )1

3 3 321 12 2

log log log log log logx xx y z z z

y y

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− + = + =⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

g = 3

log xzy

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

24. -

For the following exercises, rewrite each expression as an equivalent ratio of logs using the indicated base.

Section6.5

547

25. ( )7log 15 to base e

log log 15log =log log 7

n eb

n e

MM

b= , therefore ( ) ( )

( )7

ln 15log 15

ln 7=

26. -

For the following exercises, suppose ( )5log 6 a= and ( )5log 11 .b= Use the change-of-base formula along with properties of logarithms to rewrite each expression in terms of a and b. Show the steps for solving.

27. ( )11log 5

Since we are given ( )5log 6 a= and ( )5log 11 .b= to use we need to change this

expression with a base of 5 and realize that ( )5log 5 1= , then it follows that

( ) ( )( )

511

5

log 5 1log 5log 11 b

= =

28. -

29. 116log11⎛ ⎞⎜ ⎟⎝ ⎠

Since we are given ( )5log 6 a= and ( )5log 11 .b= to use, first we need to change this

expression with a base of 5. ( )

5

115

6log6 11log11 log 11

⎛ ⎞⎜ ⎟⎛ ⎞ ⎝ ⎠=⎜ ⎟

⎝ ⎠ . Next we need to think about rewriting

the 56log11⎛ ⎞⎜ ⎟⎝ ⎠

in terms of the given information ( )5log 11 .b= and ( )5log 6 a= Applying the

quotient rule to this yields ( ) ( )5 5 56log log 6 log 1111

a b⎛ ⎞ = − = −⎜ ⎟⎝ ⎠

, therefore it follows that

( )( ) ( )

( )

55 5

115 5

6log log 6 log 116 11log 111 log 11 log 11

a b aor

b b

⎛ ⎞⎜ ⎟ − −⎛ ⎞ ⎝ ⎠= = = −⎜ ⎟

⎝ ⎠

Numeric For the following exercises, use properties of logarithms to evaluate without using a calculator.

30. -

Section6.5

548

31. ( ) ( )( )

88

8

log 646log 2

3log 4+

First we apply the power rule rewriting ( ) ( )68 86log 2 log 2= and ( ) ( )3

8 83log 4 log 4= .

Recall rewriting each individual exponentially to solve it 1) ( )6 6 3 6

8log 2 8 2 2 2 , 2x xx written exponentially is or so x= = = = , therefore ( )68log 2 2=

2) ( ) ( )33 3 3 2 3 68log 4 8 4 2 2 2 2 , 2x x xx written exponentially is or or x= = = = = , therefore

( )38log 4 2=

3) ( ) 28log 64 8 64 8 , 2xx written exponentially is so x= = = = , therefore ( )8log 64 2=

Entering these values into the expression yields ( ) ( )( )

88

8

log 64 26log 2 2 2 1 33log 4 2

+ = + = + = .

32. -

For the following exercises, use the change-of-base formula to evaluate each expression as a quotient of natural logs. Use a calculator to approximate each to five decimal places.

33. ( )3log 22

( )3log 22log 22 log(22) log(3)log3

= = ÷ ≈ 2.81359

34. - 35. ( )6log 5.38

( )6log5.38log 5.38 log(5.38) log(6)

log6= = ÷ ≈ 0.93913

36. -

37. ( )12

log 4.7

( ) ( )12

log 4.7log 4.7 log(4.7) log(1 2)1log

2

= = ÷ ÷ ≈ 2.23266−

Extensions 38. - 39. Use the quotient rule for logarithms to find all x values such that

( ) ( )6 6log 2 log 3 1.x x+ − − = Show the steps for solving.

Section6.5

549

Using the quotient rule for logarithms we condense the left-hand side as a single logarithm:

( ) ( )6 6

6

log 2 log 3 12log 1.3

x xxx

+ − − =

+⎛ ⎞ =⎜ ⎟−⎝ ⎠

Rewriting as an exponential equation and solving for x :

( )

( )

1 26320 63

6 3203 32 6 180

3403

4

xxxx

xxx xx x

xxx

x

+=

−+

= −−

−+= −

− −+ − +

=−

−=

−=

Checking, we find that ( ) ( ) ( ) ( )6 6 6 6log 4 2 log 4 3 log 6 log 1+ − − = − is defined, so x = 4.

40. -

41. Prove that ( )( )

1loglogb

n

nb

= for any positive integers 1b > and 1.n >

Let b and n be positive integers greater than 1. Then, by the change-of-base formula,

( ) ( )( ) ( )

log 1log .log log

nb

n n

nn

b b= =

42. – Thisfileiscopyright2015,RiceUniversity.AllRightsReserved.

Section 6.6

550

Chapter 6 Exponential and Logarithmic Functions

6.6 Exponential and Logarithmic Equations Section Exercises Verbal

1. How can an exponential equation be solved?

Determine first if the equation can be rewritten so that each side uses the same base. If

so, the exponents can be set equal to each other. If the equation cannot be rewritten so

that each side uses the same base, then apply the logarithm to each side and use

properties of logarithms to solve.

2. - 3. When can the one-to-one property of logarithms be used to solve an equation? When can

it not be used? The one-to-one property can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base.

Algebraic For the following exercises, use like bases to solve the exponential equation.

4. - 5. 364 4 16x⋅ =

Find the same common base between 64, 4, and 16, which would be 4. We want to rewrite each side with the base of 4, using all the correct exponent rules. Once we have a single base expression on each side and the base is the same, we set the exponents equal and solve for x.

3

3 3 2

3 3 2

64 4 164 4 4 Rewrite each side as a power with base 4.

4 4 Product exponent rule on left side, add exponents.3 3 2 Apply the one-to-one property of exponents.

3 1 Subtrac

x

x

x

xx

+

⋅ =⋅ =

=+ =

= − t 3 from both sides.1 Divide by 3.

3x

−=

6. -

Section 6.6

551

7. 3 212 24

n n− +⋅ =

Find the same common base between 2 and 4 which would be 2. We want to rewrite each side with the base of 2, using all the correct exponent rules. Once we have a single base expression on each side and the base is the same, we set the exponents equal and solve for x.

3 2

3 22

3 2 22

3 2 2

12 2412 2 Rewrite each side as a power with base 2.2

12 2 2 Rewrite on left side as a negative exponent.2

2 2 Product exponent rule on left side, add exponents3

n n

n n

n n

n n

− +

− +

− − +

− +− +

⋅ =

⋅ =

⋅ =

=− 2 2 Apply the one-to-one property of exponents.

4 2 2 Subtract n from both sides.4 4 Add 2 to both sides.

1 Divide by 4.

n nn

nn

− = +− − =

− == − −

8. -

9. 3

22

36 21636

bb

b−=

Find the same common base between 36 and 216, which would be 6. We want to rewrite each side with the base of 6, using all the correct exponent rules. Once we have a single base expression on each side and the base is the same, we set the exponents equal and solve for b.

Section 6.6

552

( )( )

( )

32

2

3223

22

66 3

4

6 4 6 3

36 21636

66 Rewrite each side as a power with base 6.

6

6 6 Apply power to power, multiply the exponents, rule 6

6 6 Quotient exponent rule on left side, subtract expon

bb

b

bb

b

bb

b

b b b

−

−

−

− −

=

=

=

=2 6 3

ents6 6 2 6 3 Apply the one-to-one property of exponents.5 6 Add 3b to both sides.

6 Divide by 5.5

b b

b bb

b

−== −=

=

10. -

For the following exercises, use logarithms to solve.

11. 109 1x− = The terms of this exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall, since ( ) ( )log loga b= is equivalent to

a = b, we may apply logarithms with the same base on both sides of an exponential equation.

10

10

9 1 There is no easy way to get the powers to have the same base.ln 9 ln1 Take ln of both sides.

( 10) ln 9 0 Use laws of logs.ln 9 10ln 9 0

x

x

xx

−

−

=

=

− =

− = Use the distributive law.ln 9 10ln 9 Get terms containing on one side, terms without on the other.ln 9 10ln 9 Divide by ln 9.

ln 9 ln 910

x x xx

x

=

=

=

12. -

13. 10 10 42re + − = −

We want the 10re + alone on one side before we begin.

Section 6.6

553

( )

10

10

10

10 42 Add 10 to both sides32 Take the ln of both sides

ln ln 32

r

r

r

eee

+

+

+

− = −= −= −

We can stop here because ( )ln 32− is not defined, therefore there is no solution

14. -

15. 78 10 7 24p+− ⋅ − = − We want the 910 a alone on one side before we begin.

7

7

7

7

710

8 10 7 24 Add 7 to both sides.8 10 17 Divide both sides by 8

17108

17108

17log10 log Since it's in base 10, take log of b8

p

p

p

p

p

+

+

+

+

+

− ⋅ − = −

− ⋅ = − − ⋅

−=−

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

( )

( )

oth sides.

177 log10 log Use laws of logs. Remember log10 =18

177 (1) log8

177 log Subtract 7 from both sides8

17log 7 8

p

p

p

p

⎛ ⎞+ = ⎜ ⎟⎝ ⎠

⎛ ⎞+ = ⎜ ⎟⎝ ⎠

⎛ ⎞+ = ⎜ ⎟⎝ ⎠

⎛ ⎞= −⎜ ⎟⎝ ⎠

16. -

17. 9 85 8 62xe −− − = − We want the 9 8xe − alone on one side before we begin.

Section 6.6

554

( )

9 8

9 8

9 8

9 8

9 8

5 8 62 Add 8 to both sides5 54 Divide both sides by 5.

545

54 5

54ln ln Take ln of both sides.5549 8 ln ln 5

x

x

x

x

x

e

e

e

e

e

x e

−

−

−

−

−

− − = −

− = − −

−=−

=

=

⎛ ⎞− = ⎜ ⎟⎝ ⎠

( )

Use laws of logs. Remember ln e=1

549 8 (1) ln5

549 8 ln Add 8 to both sides5

549 ln 8554ln 8 5 Divide by 9 9

x

x

x

x

⎛ ⎞− = ⎜ ⎟⎝ ⎠

⎛ ⎞− = ⎜ ⎟⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞+⎜ ⎟⎝ ⎠=

18. - 19. 1 2 12 5x x+ −=

( ) ( )

( ) ( )ln 2 ln 5

ln 2 2ln 5x

− −=

−

2x+1 = 52x−1 There is no easy way to get the powers to have the same base.ln 2x+1 = ln52x−1 Take ln of both sides.

(x +1)ln 2( ) = 2x −1( )ln 5( ) Use laws of logs.

x ln 2( )+ ln 2( ) = 2x ln 5( )− ln 5( ) Use the distributive law.

x ln 2( )− 2x ln 5( ) = −ln 5( )− ln 2( ) Get terms containing x on one side, terms without x on the other.

x ln 2( )− 2ln 5( )( ) = −ln 5( )− ln 2( ) On the left hand side, factor out an x.

x = −ln 5( )− ln 2( )

ln 2( )− 2ln 5( ) Divide both sidesby ln 2( )− 2ln 5( )( )

Section 6.6

555

20. -

21. 8 87 5 95.xe + − = −

We want the 8 8xe + alone on one side before we begin. 8 8

8 8

8 8

8 8

7 5 95. Add 5 to both sides 7 90 Divide both sides by 7.

90 7

90ln ln Take ln of both sides.7

x

x

x

x

e

e

e

e

+

+

+

+

− = −

= −

−=

−⎛ ⎞= ⎜ ⎟⎝ ⎠

We can stop here because 90ln7−⎛ ⎞⎜ ⎟⎝ ⎠

is not defined, therefore there is no solution

22. -

23. 3 34 7 53.xe + − = We want the 3 3xe + alone on one side before we begin.

( )( ) ( )

3 3

3 3

3 3

3 3

3 3

4 7 53. Add 7 to both sides4 60 Divide both sides by 4.

604

15ln ln 15 Take ln of both sides.

3 3 ln ln 15 Use laws

x

x

x

x

x

e

e

e

e

e

x e

+

+

+

+

+

− =

=

=

=

=

+ =

( ) ( )( )( )( )

of logs. Remember ln e =1

3 3 (1) ln 15

3 3 ln 15 Subtract 3 from both sides

3 ln 15 3

ln 15 3 Divide by 3

3

x

x

x

x

+ =

+ =

= −

−=

24. - 25. 2 1 23 7x x+ −=

Section 6.6

556

( ) ( ) ( )( )

2 1 2

2 1 2

3 7 There is no easy way to get the powers to have the same base.ln 3 ln 7 Take ln of both sides.

(2 1) ln 3 2 ln 7 Use laws of logs.

2 ln 3

x x

x x

x x

x

+ −

+ −

=

=

+ = −

+ ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )( ) ( ) ( )

ln 3 ln 7 2ln 7 Use the distributive law.

2 ln 3 ln 7 2ln 7 ln 3 Get terms containing on one side, terms without on the other.

2 ln 3 ln 7 2ln 7 ln 3 On the left han

x

x x x x

x

= −

− = − −

− = − −

( ) ( )( ) ( )

( ) ( )( )

d side, factor out an .

2 ln 7 ln 3 Divide both sidesby 2ln 3 ln 7

2ln 3 ln 7

x

x− −

= −−

26. - 27. 3 33 6 31.xe − + = −

We want the 3 3xe − alone on one side before we begin. 3 3

3 3

3 3

3 3

3 6 31. Subtract 6 from both sides 3 37 Divide both sides by 3.

37 3

37ln ln Take ln of both sides.3

x

x

x

x

e

e

e

e

−

−

−

−

+ = −

= −

−=

−⎛ ⎞= ⎜ ⎟⎝ ⎠

We can stop here because 37ln3−⎛ ⎞⎜ ⎟⎝ ⎠

is not defined, therefore there is no solution

For the following exercises, use the definition of a logarithm to rewrite the equation as an exponential equation.

28. -

29. ( )3241log 182

=

log ( ) expbase S onent= if and only if .exponentbase S=

In this problem, the base is 324 and the exponent is 12

, therefore we rewrite this

equation 12324 18=

For the following exercises, use the definition of a logarithm to solve the equation.

30. –

Section 6.6

557

31. 98log 16x− = We want the 9log x alone on one side before we begin.

9

92

8log 16 Divide both sides by -8log 2 Rewrite exponentially9

xxx−

− == −

=

Therefore 181

x =

32. - 33. ( )2log 8 4 6 10n + + =

We want the ( )log 8 4n + alone on one side before we begin.

( )( )( )

2

2log 8 4 6 10 Subtract 6 from both sides2log 8 4 4 Divide both sides by 2

log 8 4 210 8 4 Rewrite exponentially

100 4 8 Subtract 4 from both sides96 8 Divide both sides by 812

nnn

nnn

n

+ + =

+ =

+ =

= +− =

==

Substituting 12n = into the original logarithmic functions we see the argument of the logarithm functions is positive therefore this is a solution. Therefore 12n =

34. - For the following exercises, use the one-to-one property of logarithms to solve.

35. ( ) ( )ln 10 3 ln 4x x− = −

( ) ( )ln 10 3 ln 4 10 3 4 Use the one-to-one property of the logarithm.

10 1 Add 3x to both sides10

x x

x xx

x

− = −

− = −

= −

− = Divide by 1−

Substituting 10x = − into the original logarithmic functions we see the argument of the logarithm functions is positive therefore this is a solution. Therefore 10x = −

36. -

Section 6.6

558

37. ( ) ( ) ( )log 3 log log 74x x+ − = We have to rewrite each side as a single logarithm first

( ) ( ) ( )

( )

( )

log 3 log log 743log log 74 Use the one-to-one property of the logarithm

3 74 Multiply both sides by x

3 74 Subtract x from both sides3 73 Divide by 73

373

373

x xx

xx

xx x

x

x

Therefore x

+ − =

+⎛ ⎞ =⎜ ⎟⎝ ⎠

+⎛ ⎞ =⎜ ⎟⎝ ⎠

+ ==

=

=

Substituting 373

x = into the original logarithmic functions we see the argument of the

logarithm functions is positive therefore this is a solution.

Therefore 373

x =

38. - 39. ( )4 4log 6 log 3m m− =

( )4 4log 6 log 36 3 Use the one-to-one property of the logarithm

6 4 Add m to both sides6 Divide by 4432

32

m mm m

m

m

m

Therefore m

− =− =

=

=

=

=

Substituting 32

m = into the original logarithmic functions we see the argument of the

logarithm functions is positive therefore this is a solution.

Therefore 32

m =

40. - 41. ( ) ( )2 2

9 9log 2 14 log 45n n n− = − +

Section 6.6

559

( ) ( )2 29 9

2 2

2

log 2 14 log 45

2 14 45 Use the one-to-one property of the logarithm.14 45 0 Get zero on one side before f

n n n

n n n

n n

− = − +

− = − +

− + = actoring.( 9)( 5) 0 Factor using FOIL.

9 0 or n 5 0 If a product is zero, one of the factors must be zero.9 or 5 Solve for .

n nn

x x x

− − =

− = − =

= =

Substituting 5 into the original logarithmic functions we see:

( ) ( )( ) ( )

2 29 9

9 9

log 2(5) 14(5) log 45 (5)log 20 log 20

− = − +

− = −

Because the argument of the logarithm functions is not positive, 5 is not a solution. Substituting 9 into the original logarithmic functions we see:

( ) ( )( ) ( )

2 29 9

9 9

log 2(9) 14(9) log 45 (9)log 36 log 36

− = − +

=

Because the argument of the logarithm functions is positive, 9 is a solution Therefore 9n =

42. – For the following exercises, solve each equation for x.

43. ( ) ( ) ( )log 12 log log 12x x+ = +

We have to rewrite each side as a single logarithm first ( ) ( ) ( )( ) ( )( ) ( )

log 12 log log 12log 12 log 12log 12 log 12 Use the one-to-one property of the logarithm

12 12 Subtract x from both sides12 11 Divide by 111211

x xx xx xx x

x

x

+ = +

+ =

+ =+ =

=

=

g

Substituting 1211

x = into the original logarithmic functions we see the argument of the

logarithm functions is positive therefore this is a solution.

Therefore 1211

x = .

Section 6.6

560

44. - 45. ( )2log 7 6 3x + =

The logarithm is alone on one side, so we can rewrite this in exponential form ( )2

3

log 7 6 3 rewrite exponentially2 7 6 subtract 68 7 62 7 divide by 727

xx

xx

x

+ =

= += +=

=

27

x =

Substituting 27

x = into the original logarithmic functions we see the argument of the

logarithm functions is positive therefore this is a solution. Therefore 27

x = .

46. -

47. ( ) ( ) ( )8 8 8log 6 log log 58x x+ − =

We have to rewrite each side as a single logarithm first ( ) ( ) ( )

( )

( )

8 8 8

8 8

log 6 log log 586log log 58 Use the one-to-one property of the logarithm

6 58 Multiply both sides by x

6 58 Subtract x from both sides6 57 Divide by 57

6572

19

x xx

xx

xx x

x

x

x

+ − =

+⎛ ⎞ =⎜ ⎟⎝ ⎠

+⎛ ⎞ =⎜ ⎟⎝ ⎠

+ ==

=

=

Substituting 219

x = into the original logarithmic functions we see the argument of the

logarithm functions is positive therefore this is a solution. Therefore 219

x = .

48. -

49. ( ) ( ) ( )3 3 3log 3 log 6 log 77x − =

Section 6.6

561

We have to rewrite each side as a single logarithm first ( ) ( ) ( )

( )

( )

3 3 3

3 3

log 3 log 6 log 773log log 77 Use the one-to-one property of the logarithm61 77 Multiply both sides by 22

77(2)154

xx

x

xx

− =

⎛ ⎞ =⎜ ⎟⎝ ⎠⎛ ⎞ =⎜ ⎟⎝ ⎠

==

Substituting 154x = into the original logarithmic functions we see the argument of the logarithm functions is positive therefore this is a solution. Therefore 154x = .

Graphical For the following exercises, solve the equation for x, if there is a solution. Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution.

50. -

51. ( )3log 3 2x + =

First we want the logarithm alone on one side, so we can rewrite this in exponential form ( )( )

3

31

log 3 2 Subtract 3 from both sideslog 1 rewrite exponentially3

13

xxx

x

−

+ =

= −

=

=

Substituting 13

x = into the original logarithmic functions we see the argument of the

logarithm functions is positive therefore this is a solution. Therefore 13

x = .

Section 6.6

562

52. -

53. ( )ln 5 1x − =

Since the logarithm is alone on one side already we may rewrite it exponentially; ( )

1

ln 5 1 rewrite exponentially5 add 5

5

xe xe x

− =

= −+ =

5 7.7x e= + ≈ Substituting 5 7.7x e= + ≈ into the original logarithmic functions we see the argument of the logarithm functions is positive therefore this is a solution. Therefore 5 7.7x e= + ≈ .

54. -

55. ( )37 log 4 6x− + − = −

First we want the logarithm alone on one side, so we can rewrite this in exponential form ( )( )

3

31

7 log 4 6 Add 7log 4 1

3 4 Rewrite exponentially, subtract 41 Divide by 11

1

xx

xx

xx

− + − = −

− =

= −− = − −

==

Substituting 1x = into the original logarithmic functions we see the argument of the logarithm functions is positive therefore this is a solution. Therefore 1x = .

Section 6.6

563

56. -

57. ( ) ( )log 4 2 log 4x x− = − Since there are logs on both sides we immediately can set their arguments equal

( ) ( )log 4 2 log 4 4 2 4 Use the one-to-one property of the logarithm.

2 4 Add 4x and Subtract 4 from both sides2

x x

x xxx

− = −

− = −

= −

= − Divide by 2

Substituting 2x = − into the original logarithmic functions we see the argument of the logarithm functions is positive therefore this is a solution. Therefore 2x = −

Section 6.6

564

58. - 59. ( ) ( )ln 2 9 ln 5x x+ = −

Since there are logs on both sides we immediately can set their arguments equal

( ) ( )ln 2 9 ln 5 2 9 5 Use the one-to-one property of the logarithm.

7 9 Add 5x and Subtract 9 from both sides9 Divide by 77

x x

x xx

x

+ = −

+ = −

= −

−=

Substituting 9 1.37

x = − ≈ − into the original logarithmic functions we see the argument of

the logarithm functions is positive therefore this is a solution. 9 1.37

x = − ≈ −

60. - 61. ( ) ( )2log 13 log 7 3x x+ = +

Since there are logs on both sides we immediately can set their arguments equal

( ) ( )

( )

2

2

2

log 13 log 7 3 Use the one-to-one property of the logarithm

13 7 3 Subtract 7x and 3 from both sides7 10 0 Factor into two binomials

5 ( 2) 05 2

x x

x xx xx x

x or x

+ = +

+ = +− + =

− − == =

Substituting 2x = or 5x = into the original logarithmic functions we see the argument of the logarithm functions is positive therefore they are both solutions. Therefore the solutions are 2x = and 5x = .

Section 6.6

565

62. -

63. ( ) ( ) ( )ln ln 3 ln 6x x− + =

We have to rewrite each side as a single logarithm first ( ) ( ) ( )

( )

( )

ln ln 3 ln 6

ln ln 6 Use the one-to-one property of the logarithm3

6 Multiply both sides by x+33

6 36 18 Subtract 6x from both sides

5 1818 Divide by 5

53.6

x xx

xx

xx xx xx

x

x

− + =

⎛ ⎞ =⎜ ⎟+⎝ ⎠

=+

= += +

− =

= −−

= −

There is a solution for x , however when 3.6x = − is substituted into the original equation because the argument of the logarithm functions is not positive, there is no solution .

Section 6.6

566

For the following exercises, solve for the indicated value, and graph the situation showing the solution point.

64. -

65. The formula for measuring sound intensity in decibels D is defined by the equation

0

10log ,ID

I⎛ ⎞

= ⎜ ⎟⎝ ⎠

where I is the intensity of the sound in watts per square meter and

120 10I −= is the lowest level of sound that the average person can hear. How many

decibels are emitted from a jet plane with a sound intensity of 28.3 10⋅ watts per square meter? Substituting the given information into

( ) ( )( )( ) ( )( )

( )( )( )

214

120

14

8.3 1010log 10log 10log(8.3 10 )10

10 log 10 log 8.3

10 14log 10 log 8.310 14 log 8.310 14 0.9191149.2

ID

I −

⎛ ⎞ ⎛ ⎞⋅= = = ⋅⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠= +

= +

= +

= +≈

Therefore about 149.2 decibels.

66. -

Section 6.6

567

Technology For the following exercises, solve each equation by rewriting the exponential expression using the indicated logarithm. Then use a calculator to approximate x to 3 decimal places.

67. ( )1000 1.03 5000t= using the common log.

( )( ) ( )

( )( )( ) ( )

1000 1.03 5000

1000 1.03 5000 1.03 alone on one side1000 1000

1.03 5 take log of both sides to solve for t

log 1.03 log5 recall power rule of logs on left side log 1.03 log5 Divide by log 1.03

log(5)log(

t

tt

t

t

get

t

t

=

=

=

=

=

=1.03)

log(5) log(1.03) 54.44868501 54.449t = ÷ = ≈

68. -

69. ( )33 1.04 8t= using the common log

( )( ) ( )

( )

( )

( ) ( )

3

33

3

3

3 1.04 8

3 1.04 8 1.04 alone on one side3 3

81.04 take log of both sides to solve for t3

8log 1.04 log recall power rule of logs on left side 383 log 1.04 log Divide by 3log 1.043

8log3

t

tt

t

t

get

t

t

=

=

=

⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛

=3log(1.04)log(8 3) (3log(1.04)) 8.3359801 8.336t

⎞⎜ ⎟⎝ ⎠

= ÷ ÷ = ≈

70. -

Section 6.6

568

71. 0.1250 10te− = using the natural log

( )

0.12 0.12

0.12

0.12

0.12

50 10 alone on one side50 10 20,000 1650 hit math Frac to get a fraction instead of a repeating decimal

50 501 take ln of both sides to solve for t5

1ln ln reca5

t t

t

t

t

e get ee

e

e

− −

−

−

−

=

= ÷

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

( )

ll power rule of logs on leftt side and ln e = 1

10.12 ln ln510.12 ln Divide by 0.1251ln5

0.12ln(1 5) 0.12 13.4119826 13.412

t e

t

t

t

⎛ ⎞− = ⎜ ⎟⎝ ⎠⎛ ⎞− = −⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟⎝ ⎠=

−= ÷ ÷ − = ≈

For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth.

72. -

73. ( ) ( )ln 3 ln 4.4 6.8 2x+ + =

First we want the logarithm alone on one side, so we can rewrite this in exponential form

( ) ( )( )( )

( )2

2

2

ln 3 ln 4.4 6.8 2ln 3 4.4 6.8 2 Adding individual logs is product rule as single logln 13.2 20.4 2

13.2 20.4 Rewrite exponentially, subtract 20.420.4 13.220.4 Divide by 13.2

13.2( ^ (2) 20.4) 1

xxx

e xe xe

x

x e

+ + =

+ =

+ =

= +− =−

=

= − ÷ 3.2 0.9856775683 .9857= − ≈ −

0.9857x ≈ −

74. - 75. Atmospheric pressure P in pounds per square inch is represented by the formula

0.2114.7 ,xP e−= where x is the number of miles above sea level. To the nearest foot, how high is the peak of a mountain with an atmospheric pressure of 8.369 pounds per square inch? (Hint: there are 5280 feet in a mile) We are being asked to solve for x when 8.369P =

Section 6.6

569

( ) ( )

0.21

0.21 0.21

0.21

0.21

0.21

14.78.369 14.7 alone on one side8.36914.7

.56913197279 take ln of both sides to solve for xln .56913197279 ln recall power rule of logs on right side and l

x

x x

x

x

x

P ee get e

e

ee

−

− −

−

−

−

==

=

=

=

( ) ( )( )( )

( )

n e = 1ln .56913197279 0.21 lnln .56913197279 0.21 Divide by 0.21ln .56913197279

0.21ln .56913197279 0.21 2.682443289

x ex

x

x miles

= −

= − −

=−

= ÷ − =

= about

Since 1 mile is 5280 feet, we take 2.682443289(5280) 14163.30057= . Therefore The peak of this mountain would be about 14,163 feet above sea level

76. - Extensions

77. Use the definition of a logarithm along with the one-to-one property of logarithms to prove that log .xbb x= Let log .xbb n= We show that .n x= When rewriting exponentbase n= as a logarithm it would be log exponentbase n = . Therefore it follows if I write log xbb n= as a logarithm it would be

log log .b bn x= By the one-to-one property of logarithms, .n x= Therefore, log .xbb x= 78. -

Section 6.6

570

79. Recall the compound interest formula 1 .ktr

A ak

⎛ ⎞= +⎜ ⎟⎝ ⎠

Use the definition of a logarithm

along with properties of logarithms to solve the formula for time t.

1 . 1 alone on one side

1 take ln of both sides to solve for t

ln ln 1 recall power rule of logs on right side and ln e = 1

ln ln

kt kt

kt

kt

r rA a get

k kA ra k

A ra k

Akt

a

⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= +⎜ ⎟⎝ ⎠⎛ ⎞⎛ ⎞ ⎛ ⎞= +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞ =⎜ ⎟⎝ ⎠

1 Divide by ln 1

ln

ln 1

ln ln

ln 1 ln 1k

r rk

k kAa t

rkk

A Aa at

r rkk k

⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠ =⎛ ⎞+⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= =⎛ ⎞ ⎛ ⎞⎛ ⎞+⎜ ⎟ +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

80. - Thisfileiscopyright2015,RiceUniversity.AllRightsReserved.

Section 6.7

571

Chapter 6 Exponential and Logarithmic Functions 6.7 Exponential and Logarithmic Models

Section Exercises Verbal

1. With what kind of exponential model would half-life be associated? What role does half-life play in these models? Half-life is a measure of decay and is thus associated with exponential decay models. The half-life of a substance or quantity is the amount of time it takes for half of the initial amount of that substance or quantity to decay.

2. - 3. With what kind of exponential model would doubling time be associated? What role does

doubling time play in these models? Doubling time is a measure of growth and is thus associated with exponential growth models. The doubling time of a substance or quantity is the amount of time it takes for the initial amount of that substance or quantity to double in size.

4. - 5. What is an order of magnitude? Why are orders of magnitude useful? Give an example to

explain. An order of magnitude is the nearest power of ten by which a quantity exponentially grows. It is also an approximate position on a logarithmic scale; Sample response: Orders of magnitude are useful when making comparisons between numbers that differ by a great amount. For example, the mass of Saturn is 95 times greater than the mass of Earth. This is the same as saying that the mass of Saturn is about 210 times, or 2 orders of magnitude greater, than the mass of Earth.

Numeric

6. -

For the following exercises, use the logistic growth model f (x) =150

1+ 8e−2x.

7. Find and interpret f (0). Round to the nearest tenth.

2 2(0) 0150 150 150 150 150( ) (0) 16.7

1 8 91 8 1 8 1 8xf x fe e e− −

= ⇒ = = = = ≈++ + +

f (0) ≈16.7; (0)f represents

the amount initially (when t = 0) present is about 16.7 units. 8. - 9. Find the carrying capacity.

Since the equation is in the form 2

150( ) , ( ) , 1501 1 8bx x

cf x given f x c

ae e− −= = =

+ +, which is

the carrying capacity = 150

Section 6.7

572

10. -

11. Determine whether the data from the table could best be represented as a function that is

linear, exponential, or logarithmic. Then write a formula for a model that represents the data.

x –2 –1 0 1 2 3 4 5

f(x) 0.694 0.833 1 1.2 1.44 1.728 2.074 2.488

If you take any f(x) value and divide it by the preceding one, .833 .694 1 .833 1.2 1÷ = ÷ = ÷ tells us that 1.2 is the ratio increase which makes this model exponential with a base of 1.2. Therefore exponential; ( ) 1.2xf x =

12. - Technology For the following exercises, enter the data from each table into a graphing calculator and graph the resulting scatter plots. Determine whether the data from the table could represent a function that is linear, exponential, or logarithmic.

13. x 1 2 3 4 5 6 7 8 9 10

f(x) 2 4.079 5.296 6.159 6.828 7.375 7.838 8.238 8.592 8.908

Using a graphing calculator, you can plot individual points using STAT Plot. Press 2nd STAT PLOT, which is above Y=. Highlight 1:Plot1…On, Enter, Choose what type of mark you’d like , enter. To enter coordinates hit the STAT button, 1:Edit, enter. It brings L1 and L2 and L3. Enter the x values down column L1 and then the corresponding y-values in L2 column. Once they are all in exit that screen and hit GRAPH. It should plot the points for you. See below. Because of the nature of this curve (being concave down), it is logarithmic.

Section 6.7

573

14. -

15. x 4 5 6 7 8 9 10 11 12 13

f(x) 9.429 9.972 10.415 10.79 11.115 11.401 11.657 11.889 12.101 12.295

Using a graphing calculator, you can plot individual points using STAT Plot. Press 2nd STAT PLOT, which is above Y=. Highlight 1:Plot1…On, Enter, Choose what type of mark you’d like , enter. To enter coordinates hit the STAT button, 1:Edit, enter. It brings L1 and L2 and L3. Enter the x values down column L1 and then the corresponding y-values in L2 column. Once they are all in exit that screen and hit GRAPH. It should plot the points for you. See below. Because of the nature of this curve (being concave down), it is logarithmic.

Section 6.7

574

16. - For the following exercises, use a graphing calculator and this scenario: the population of a fish

farm in t years is modeled by the equation ( ) 0.6

1000 .1 9 tP t

e−=

+

Graph the function. Since we know that the highest y value will be 1000 (the carrying capacity) , we should make our window to include that. A nice window to show the graph would be

1

min 20, max 20, 2, min 100, max 1100, 100Y 1000 (1 9 ^ ( 0.6 )),

x x xscl y y ysclEnter e x graph

= − = = − = =

= ÷ + −

17.

Section 6.7

575

18. - 19. To the nearest tenth, what is the doubling time for the fish population?

Doubling time follows the model given in the Chapter

0ln 2, doubling time is , where k>0ktA A e tk

= = . Therefore since

ln 2ln ln.6, 1.3569 1.4ln 0.6

k b t= − = − = = ≈−

which is about 1.4 years

20. - 21. To the nearest tenth, how long will it take for the population to reach 900?

We use the original model and solve for t, plugging in 900 for the population size

Section 6.7

576

( )

( )

0.6

0.6

0.6 0.6

0.6

0.6

0.6

0.6

10001 9

10009001 9

900 1 9 1000 multiply both sides by 1 9

900 8100 1000 distribute, subtract 9008100 100 divide by 8100

1= take the ln of both sides81

1ln =ln

t

t

t t

t

t

t

t

P te

ee e

ee

e

e

−

−

− −

−

−

−

−

=+

=+

+ = +

+ ==

log properties and lne = 1811.6 ln ln811.6 ln divide by .6811ln81 (ln(1 81)) ( .6) 7.324081924 7.3.6

t e

t

t

⎛ ⎞⎜ ⎟⎝ ⎠⎛ ⎞− = ⎜ ⎟⎝ ⎠⎛ ⎞− = −⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟⎝ ⎠= = ÷ ÷ − = ≈−

It would take about 7.3 years.

22. - Extensions

23. A substance has a half-life of 2.045 minutes. If the initial amount of the substance was 132.8 grams, how many half-lives will have passed before the substance decays to 8.3 grams? What is the total time of decay? If we take 132.8 and repeatedly divide it by 2, we see it takes 4 cycles of the half life to reach a size of 8.3. Therefore since the half-life is 2.045 minutes we multiply 4*2.045 8.18= . Four half-lives; 8.18 minutes

24. -

25. Recall the formula for calculating the magnitude of an earthquake, 0

2 log .3

SM

S⎛ ⎞

= ⎜ ⎟⎝ ⎠

Show

each step for solving this equation algebraically for the seismic moment .S

Section 6.7

577

0

03

20

03

20

32

0

2 3log multiply both sides by 3 2

3 log rewrite this exponentially, base 102

10 multiply by S

10

10

M

M

M

SM

S

SM

S

SS

S S

S S

⎛ ⎞= ⎜ ⎟

⎝ ⎠⎛ ⎞

= ⎜ ⎟⎝ ⎠

=

=

=

26. -

27. Prove that ( )lnx bxb e= for positive 1.b ≠

Let xy b= for some non-negative real number b such that 1.b ≠ Then,

( ) ( )( ) ( )

( ) ( )

( )

ln ln

ln

ln lnln ln

x

y x b

x b

y b

y x b

e ey e

=

=

=

= Real-World Applications For the following exercises, use this scenario: A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour.

28. - 29. Write an exponential model representing the amount of the drug remaining in the

patient’s system after t hours. Then use the formula to find the amount of the drug that would remain in the patient’s system after 3 hours. Round to the nearest milligram. Using the information we found ( ) ( )ln ln.7125b x xy ae e= = . So the exponential model would

be A = 125e −0.3567t( ); Using this model we will plug in t = 3 hours. ( )ln.7 3(3) 125 42.875 43f e mg= = ≈ the amount of drug remaining is 43A ≈ mg

30. - For the following exercises, use this scenario: A tumor is injected with 0.5 grams of Iodine-125, which has a decay rate of 1.15% per day.

31. To the nearest day, how long will it take for half of the Iodine-125 to decay? Since this has a decay rate of 1.1%, that means that 100% - 1.15% = 98.85% is remaining. We are being asked for the half-life, so since ln ln .9985k b= = , the half-life

is( )

ln 2 ln 2 59.92642602 60.ln .9885

tk

− −= = = ≈ About 60 days

32. -

Section 6.7

578

33. A scientist begins with 250 grams of a radioactive substance. After 225 minutes, the sample has decayed to 32 grams. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest minute, what is the half-life of this substance? Since the initial amount 0 250A g= and after 225 minutes it decays to 32 grams we need to find k, so plugging in the given information we see:

0 0

(225min)

225

32 250 divide both sides by 250 to get e0.128 take the ln of both sides and use power rule on rigft side

ln 0.128 225 ln ln e = 1ln 0.128 225 divide both sides b

kt

k kt

k

A A e

e alone

ek ek

=

=

=

=

=

0.00914

y 225ln 0.128

225.0091365556 .00914

( ) 250 t

k

k

Therefore f t e−

=

= − ≈ −

=

To find the half-life, ln 2 ln 2 75.8366 76.0.00914

tk

− −= = = ≈

−

( 0.00914 )( ) 250 ;tf t e −= The half-life is about 76 minutes

34. -

35. The half-life of Erbium-165 is 10.4 hours. What is the hourly decay rate? Express the decimal result to four significant digits and the percentage to two significant digits.

Since ln 2t

k−

= , it follows that ln 2 ln 2 0.0666487674 0.0666510.4

kt

− −= = = − ≈ − .

0.0667,r ≈ − So the hourly decay rate is about 6.67%

36. -

37. A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was 1350 bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 hours?

Section 6.7

579

Using the given information we will solve for k, the growth constant and then we will use the model, to find f(3). Doubling the initial amount is 02A .

0 0(20)

0 0 020

2 divide both sides by

2 take the ln of both sides and use power rule on rigft sideln 2 20 ln ln e = 1ln 2 20 divide both sides by 20ln 220

.034657359( )

kt

k

k

A A e

A A e A

ek ek

k

k

Therefore f t

=

=

=

=

=

=

≈0.0346573591350 te=

Plugging 3 hours into this we must change it to minutes, therefore t = 3 hours means 60*3=180 minutes. Inputting this into this model yields

0.034657359(180)(180) 1350 691199.9965f e= ≈ . Thus the model, rounding to five significant

digits is (0.03466 )( ) 1350 ;tf t e= and after 3 hours, which is 180 minutes : (180) 691,200P ≈

For the following exercises, use this scenario: A biologist recorded a count of 360 bacteria present in a culture after 5 minutes and 1000 bacteria present after 20 minutes. 38. - 39. Rounding to six significant digits, write an exponential equation representing this

situation. To the nearest minute, how long did it take the population to double? The equation was shown in the previous, rounding the rate to six significant digits is

(0.068110 )( ) 256 ;tf t e= doubling time is ln 2 ln 2 10.1768 10.0681100832

tk

= = = ≈ , to the nearest

minute is about 10 minutes For the following exercises, use this scenario: A pot of boiling soup with an internal temperature of 100° Fahrenheit was taken off the stove to cool in a 69 F° room. After fifteen minutes, the internal temperature of the soup was 95 F.°

40. - 41. To the nearest minute, how long will it take the soup to cool to 80 F?°

Now we use our model and plug in T(t) = 70 and solve for t

Section 6.7

580

( 0.011726 )

( 0.011726 )

( 0.011726 )

0.011726

( ) 31 6980 31 6911 3131 3111 Take the ln of both sides, power rule on right side and ln e = 131

11ln 0.011726 divide by 0.01172631

11ln31

t

t

t

t

T t eee

e

t

−

−

−

−

= += +

=

=

⎛ ⎞ = − −⎜ ⎟⎝ ⎠

⎛

0.01172688.3585137

t

t

⎞⎜ ⎟⎝ ⎠ =

−=

It will cool to 80 degrees in about 88 minutes. 42. -

For the following exercises, use this scenario: A turkey is taken out of the oven with an internal temperature of 165° Fahrenheit and is allowed to cool in a 75 F° room. After half an hour, the internal temperature of the turkey is 145 F.°

43. Write a formula that models this situation.

The model formula we use is ( ) ktsT t Ae T= + and we know 75sT = ° and

165 75 90A = − = , therefore to find k, we plug in the fact that when 30min, (30) 145t T= =

(30)

(30)

30

( )

145 90 75 Subtract 75 and then divide by 9070 9090 907 Take the ln of both sides, power rule on right side and ln e = 19

7ln 30 9

kts

k

k

k

T t Ae T

e

e

k

= +

= +

=

=

⎛ ⎞ =⎜ ⎟⎝ ⎠

divide by 30

7ln9

300.0083771476

k

k

⎛ ⎞⎜ ⎟⎝ ⎠ =

≈ −

Therefore our model would be: ( 0.008377 )( ) 90 75,tT t e −= + where t is in minutes.

Section 6.7

581

44. - 45. To the nearest minute, how long will it take the turkey to cool to 110 F?°

To solve this we use our model and T(t)=110, and solve for t. ( 0.008377 )

( 0.008377 )

( 0.008377 )

0.008377

( ) 90 75110 90 75 subtract 75, divide by 9035 9090 907 Take the ln of both sides, power rule on right side and ln e = 1

187ln 0.008377

18

t

t

t

t

T t eee

e

t

−

−

−

−

= += +

=

=

⎛ ⎞ = −⎜ ⎟⎝ ⎠

divide by 0.008377

7ln18

0.008377112.7446

t

ttt

−

⎛ ⎞⎜ ⎟⎝ ⎠ =

−=

for the turkey to cool to 110 degrees it will take about 113 minutes.

For the following exercises, find the value of the number shown on each logarithmic scale. Round all answers to the nearest thousandth.

46. -

47.

We solve the equation where ( )log 1.5x = . To solve this we rewrite it exponentially.

10

1.5

log 1.5

1031.623

x

xx

=

=

≈

48. -

49. Recall the formula for calculating the magnitude of an earthquake, 0

2 log .3

SM

S⎛ ⎞

= ⎜ ⎟⎝ ⎠

One

earthquake has magnitude 3.9 on the MMS scale. If a second earthquake has 750 times

Section 6.7

582

as much energy as the first, find the magnitude of the second quake. Round to the nearest hundredth.

11

0

2 12

0 0

1 1 12

0 0 0

12

0

23.9 log3

7502 2log log3 3

7502 2 2 2log log 750 log(750) log3 3 3 3

2 2log(750) log 1.9167 3.9 5.8167 5.823 3

SM

S

S SM

S S

S S SM

S S S

SM

S

⎛ ⎞= = ⎜ ⎟

⎝ ⎠

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = ⋅ = +⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞= + = + = ≈⎜ ⎟

⎝ ⎠

For the following exercises, use this scenario: The equation ( ) 0.7

5001 49 tN t

e−=

+ models the

number of people in a town who have heard a rumor after t days. 50. - 51. To the nearest whole number, how many people will have heard the rumor after 3 days?

To solve this we plug in t = 3.

( ) 0.7(3)

5003 500 (1 49* ^ ( 0.7*3)) 71.42484729 711 49

N ee−

= = ÷ + − = ≈+

. (3) 71N ≈

52. -

53. A doctor injects a patient with 13 milligrams of radioactive dye that decays exponentially. After 12 minutes, there are 4.75 milligrams of dye remaining in the patient’s system. Which is an appropriate model for this situation?

A. ( ) ( )13 0.0805 tf t =

B. ( ) 0.919513 tf t e=

C. ( 0.0839 )( ) 13 tf t e −=

D. ( ) 0.83925

4.751 13 tf t

e−=

+

To answer this we must find k, the rate by plugging the given information in

Section 6.7

583

0(12)

12

( )

4.75 13 divide both sides by 13 to get e.3653846154 take the ln of both sides and use power rule on rigft side

ln .3653846154 12 ln ln e = 1ln .3653846154 12 divide

kt

k kt

k

A t A e

e alone

ek ek

=

=

=

=

=

0.0839003949

both sides by 5730ln .3653846154

120.0839003949

( ) 13 t

k

k

Therefore f t e−

=

≈ −

=

Choice C

Thisfileiscopyright2015,RiceUniversity.AllRightsReserved.

Section 6.8

584

Chapter 6 Exponential and Logarithmic Functions 6.8 Fitting Exponential Models to Data

Section Exercises Verbal

1. What situations are best modeled by a logistic equation? Give an example, and state a case for why the example is a good fit. Logistic models are best used for situations that have limited values. For example, populations cannot grow indefinitely since resources such as food, water, and space are limited, so a logistic model best describes populations.

2. - 3. What is regression analysis? Describe the process of performing regression analysis on

a graphing utility. Regression analysis is the process of finding an equation that best fits a given set of data points. To perform a regression analysis on a graphing utility, first list the given points using the STAT then EDIT menu. Next graph the scatter plot using the STAT PLOT feature. The shape of the data points on the scatter graph can help determine which regression feature to use. Once this is determined, select the appropriate regression analysis command from the STAT then CALC menu.

4. - 5. What does the y-intercept on the graph of a logistic equation correspond to for a

population modeled by that equation? The y-intercept on the graph of a logistic equation corresponds to the initial population for the population model.

Graphical

For the following exercises, match each scatterplot with the function of best fit.

Section 6.8

585

6. - 7. 5.598 1.912ln( )y x= −

Because this is a logarithmic function but with a negative sign in front of the function which reflects it about the x-axis its curve would then be either choice C or E, then we look at the point (1,5.598) should be on the plots, therefore function Choice = C

8. - 9. 4.607 2.733ln( )y x= +

Because this is a logarithmic function it would be between Choice B or D, and since the point (1, 4.607) should be on the graph it would be Choice = B

10. - Numeric

11. To the nearest whole number, what is the initial value of a population modeled by the

logistic equation0.68

175( ) ?1 6.995 tP t

e−=

+ What is the carrying capacity?

The initial value means when t = 0, thus 0.68(0)175 175(0) 21.88868043

7.9951 6.995P

e−= = =

+,

rounded to the nearest whole number (0) 22P = ; The carrying capacity is the numerator of the logistic growth model expression , thus it = 175 .

12. - 13. A logarithmic model is given by the equation ( )( ) 67.682 5.792ln .h p p= − To the

nearest hundredth, for what value of p does ( ) 62?h p = We will solve the given function for p:

Section 6.8

586

( )( )( )

( )0.9810082873

62 67.682 5.792ln62 67.682 5.792ln

5.682 5.792ln divide by 5.7920.9810082873 ln rewrite exponentially

2.667144134

ppp

pe p

p

= −

− = −

− = − −

=

==

2.67p ≈ 14. - 15. What is the y-intercept on the graph of the logistic model given in Exercise 14?

to find the y-intercept we plug x = 0 into the function

0.42(0)90 90(0) 15

61 5P

e−= = =

+. Therefore the y-intercept is ( )0, 15

Technology For the following exercises, use this scenario: The population P of a koi pond over x months

is modeled by the function 0.28

68( ) .1 16 xP x

e−=

+

16. - 17. What was the initial population of koi?

Algebraically, the initial value means when t = 0, thus 0.28(0)68 68(0) . 4

171 16P

e−= = =

+,

Using the graph in your graphing calculator hit 2nd calc, for calculate, then 1:value, enter, your display will be x=, input 0 (zero), enter. Read the display at the bottom x=0 y =4. answer: 4 koi

18. - 19. How many months will it take before there are 20 koi in the pond?

Enter a second function into the graph 2Y 20= . Then find the point of intersection of these two graphs. Recall , 2nd CALC, 5:intersection, move cursor to the point of intersection and hit enter 3 times. Read the display at the bottom x= 6.7754285 y =20. Therefore it will take about 6.8 months.

20. -

Section 6.8

587

For the following exercises, use this scenario: The population P of an endangered species

habitat for wolves is modeled by the function 0.462

558( ) ,1 54.8 xP x

e−=

+ where x is given in years.

21. Graph the population model to show the population over a span of 10 years. In to your graphing calculator enter the following for Y1

( )( )1Y 558 1 54.8 ^ .462e x= ÷ + − . You want your y-values to go to 558 and your data

is in terms of years, so 10 years for the x-values to reach, a nice window to show the graph would be min 0, max 20, min 0, max 600,scl50x x y y= = = =

22. - 23. How many wolves will the habitat have after 3 years?

Using the graph in your graphing calculator hit 2nd calc, for calculate, then 1:value, enter, your display will be x=, input 3, enter. Read the display at the bottom x= 3 y =37.948771. (3) 37.9,P ≈ so about 38 wolves

24. - 25. Use the intersect feature to approximate the number of years it will take before the

population of the habitat reaches half its carrying capacity. Half its carrying capacity would be one half of 558, which is 279. Enter a second function into the graph 2Y 279= . Then find the point of intersection of these two

Section 6.8

588

graphs. Recall , 2nd CALC, 5:intersection, move cursor to the point of intersection and hit enter 3 times. Read the display at the bottom x=8.6659961 y = 279 It will take approximately 8.7 years for the population to reach half capacity.

For the following exercises, use the table of data shown.

x 1 2 3 4 5 6

f(x) 1125 1495 2310 3294 4650 6361

26. -

27. Use the regression feature to find an exponential function that best fits the data in the table. To Find the equation that models the data. Select “ExpReg” from the STAT then CALC menu. Use the values returned for a and b to record the model, .xy ab= Your screen will display

Section 6.8

589

* ^776.68242921.426021377

y a b xab

=

=

=

Therefore the exponential function that best fits the data in the table is

( )( ) 776.682 1.426 xf x =

28. - 29. Graph the exponential equation on the scatter diagram.

Input ( )1Y 776.682 ^ 0.3549e x= .

30. -

For the following exercises, use the table of data shown.

x 1 2 3 4 5 6

f(x) 555 383 307 210 158 122

31. Use a graphing calculator to create a scatter diagram of the data.

Using a graphing calculator, you can plot individual points using STAT Plot. Press 2nd STAT PLOT, which is above Y=. Highlight 1:Plot1…On, Enter, Choose what type of mark you’d like , enter. To enter coordinates hit the

Section 6.8

590

STAT button, 1:Edit, enter. It brings L1 and L2 and L3. Enter the x values down column L1 and then the corresponding y-values in L2 column. Once they are all in exit that screen and hit GRAPH. It should plot the points for you. You have to go to the y= menu and make sure you highlight Plot1 above the Y1 symbol. Also Use ZOOM [9] to adjust axes to fit the data.

32. - 33. Write the exponential function as an exponential equation with base e.

Since ( ) ( )ln ln0.738 . .303811454731.92 731.92b x x xy ae e e −= = = . Therefore 0.3038( ) 731.92 xf x e−=

34. -

35. Use the intersect feature to find the value of x for which ( ) 250.f x = Enter a second function into the graph 2Y 250= . Then find the point of intersection of these two graphs. Recall , 2nd CALC, 5:intersection, move cursor to the point of intersection and hit enter 3 times. Read the display at the bottom x= 3.5359128 y =250. Therefore when ( ) 250,f x = 3.5.x ≈

Section 6.8

591

For the following exercises, use the table of data shown.

x 1 2 3 4 5 6

f(x) 5.1 6.3 7.3 7.7 8.1 8.6

36. - 37. Use the LOGarithm option of the REGression feature to find a logarithmic function of

the form ( )lny a b x= + that best fits the data in the table.

To Find the logarithmic equation that models the data. Select “LnReg” from the STAT then CALC menu. Use the values returned for a and b to record the model,

( )lny a b x= + Your screen will display

ln5.0627585491.933874889

y a b xab

= +

=

=

Therefore the logarithmic function that best fits the data in the table is ( )( ) 5.063 1.934lnf x x= +

38. -

Section 6.8

592

39. Graph the logarithmic equation on the scatter diagram. You have done this in the previous exercise, enter ( )1Y 5.063 1.934ln x= +

40. -

For the following exercises, use the table of data shown:

x 1 2 3 4 5 6 7 8

f(x) 7.5 6 5.2 4.3 3.9 3.4 3.1 2.9

41. Use a graphing calculator to create a scatter diagram of the data.

So Using a graphing calculator, you can plot individual points using STAT Plot. Press 2nd STAT PLOT, which is above Y=. Highlight 1:Plot1…On, Enter, Choose what type of mark you’d like , enter. To enter coordinates hit the STAT button, 1:Edit, enter. It brings L1 and L2 and L3. Enter the x values down column L1 and then the corresponding y-values in L2 column. Once they are all in exit that screen and hit GRAPH. It should plot the points for you. You have to go to the y= menu and make sure you highlight Plot1 above the Y1 symbol. Also Use ZOOM [9] to adjust axes to fit the data.

Section 6.8

593

42. - 43. Use the logarithmic function to find the value of the function when 10.x =

Input 1Y 7.544 2.268ln( )x= − , and 2nd CALC value x= , input 10,enter. Your display will say ERROR. This is because we used ZOOMSTAT and x= 10 is NOT in the domain of our graph. You need to go to the window and change xmax to 10, then it will read your value for you. Display x = 10 y = 2.321737, therefore (10) 2.3f ≈ .

44. -

45. Use the intersect feature to find the value of x for which ( ) 8.f x = Enter a second function into the graph 2Y 8= . Then find the point of intersection of these two graphs. Recall , 2nd CALC, 5:intersection, move cursor to the point of intersection and hit enter 3 times. Read the display at the bottom x= .81786483 y =8. When ( ) 8,f x = 0.82.x ≈

For the following exercises, use the table of data shown.

Section 6.8

594

x 1 2 3 4 5 6 7 8 9 10

f(x) 8.7 12.3 15.4 18.5 20.7 22.5 23.3 24 24.6 24.8

46. -

47. Use the LOGISTIC regression option to find a logistic growth model of the form

1 bx

cy

ae−=

+that best fits the data in the table.

To find the LOGISTIC equation that models the data. Select “B:Logistic” from the STAT then CALC menu. Use the values returned for a , b and c to record the model,

1 bx

cy

ae−=

+

Your screen will display / (1 ^ ( ))

3.181871105.545333981925.08133836

y c ae bxabc

= + −

=

=

=

Therefore the LOGISTIC function that best fits the data in the table is

0.545

25.081( )1 3.182 xf x

e−=

+

48. -

49. To the nearest whole number, what is the predicted carrying capacity of the model? The numerator of the model is the carrying capacity therefore it is about 25 .

50. - For the following exercises, use the table of data shown.

Section 6.8

595

x 0 2 4 5 7 8 10 11 15 17

( )f x 12 28.6 52.8 70.3 99.9 112.5 125.8 127.9 135.1 135.9

51. Use a graphing calculator to create a scatter diagram of the data.

Using a graphing calculator, you can plot individual points using STAT Plot. Press 2nd STAT PLOT, which is above Y=. Highlight 1:Plot1…On, Enter, Choose what type of mark you’d like , enter. To enter coordinates hit the STAT button, 1:Edit, enter. It brings L1 and L2 and L3. Enter the x values down column L1 and then the corresponding y-values in L2 column. Once they are all in exit that screen and hit GRAPH. It should plot the points for you. You have to go to the y= menu and make sure you highlight Plot1 above the Y1 symbol. Also Use ZOOM [9] to adjust axes to fit the data.

52. -

Section 6.8

596

53. Graph the logistic equation on the scatter diagram. Input ( )( )1Y 136.068 1 10.324 ^ 0.480e x= ÷ + −

54. - 55. Use the intersect feature to find the value of x for which the model reaches half its

carrying capacity. Half its carrying capacity would be one half of 136, which is 68. Enter a second function into the graph 2Y 68= . Then find the point of intersection of these two graphs. Recall , 2nd CALC, 5:intersection, move cursor to the point of intersection and hit enter 3 times. Read the display at the bottom x=4.8613995 y = 68 When ( ) 68,f x = 4.9.x ≈

Section 6.8

597

Extensions

56. - 57. Use a graphing utility to find an exponential regression formula ( )f x and a logarithmic

regression formula ( )g x for the points ( )1.5, 1.5 and ( )8.5, 8.5 . Round all numbers to 6 decimal places. Graph the points and both formulas along with the line y x= on the same axis. Make a conjecture about the relationship of the regression formulas. To find an exponential regression formula ( )f x exponential regression formula: We input the two points x values and y values in for STAT L1 and L2, hit “ExpReg” in the StatCalc menu and obtain a and b putting them in the form .xy ab=

f (x) = 1.034341 1.281204( )x;

To find a logarithmic function of the form ( )lny a b x= + that best fits the data in the table. Select “LnReg” from the STAT then CALC menu. Use the values returned for a and b to record the model to obtain the logarithmic regression formula: g(x) = 4.035510ln(x) − 0.136259; The regression curves are symmetrical about ,y x= so it appears that they are inverse functions.

58. -

Section 6.8

598

59. Find the inverse function ( )1f x−

for the logistic function ( ) .1 bx

cf x

ae−=

+ Show all

steps.

Let .1 bx

cy

ae−=

+ Solving for x we get

1

1

1

1

bx

bx

bx

bx

cy

aec

aeyc

aeycyea

−

−

−

−

=+

+ =

= −

−=

Taking the natural log of each side:

( )

( )

( )

( )

1ln ln

1ln writing right hand sides as individual logs, division is subtraction

1ln 1 ln multiply by

ln 1 ln

ln ln 1

bx

cyea

cybxa

cbx a

y bc ay

xb

cay

xb

−

⎛ ⎞−⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞−⎜ ⎟⎜ ⎟− =⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞ −

− = − −⎜ ⎟⎝ ⎠⎛ ⎞

− − +⎜ ⎟⎝ ⎠=

⎛ ⎞− −⎜ ⎟

⎝ ⎠=

Therefore, ( )( )

1ln ln 1c

axf x

b−

⎛ ⎞− −⎜ ⎟⎝ ⎠=

60. -

Chapter 6 Review Exercises

599

Chapter 6 Review Exercises Section 6.1

1. Determine whether the function ( )156 0.825 ty = represents exponential growth, exponential decay, or neither. Explain If the base <1, this base=.825, it is exponential decay; The growth factor, 0.825, is between 0 and 1.

2. - 3. Find an exponential equation that passes through the points ( )2, 2.25 and (5, 60.75).

Because we don’t have the initial value (when x = 0) we substitute both points into an equation of the form ( ) xf x ab= , and then solve the system for a and b. Substituting

( )2, 2.25 gives a first equation of 22.25 ab= substituting (5, 60.75).gives a second

equation of 560.75 ab= . Solving the first equation for a in terms of b we divide both

sides by 2b :2

2 2

2.25 abb b

= , therefore 2

2.25a

b= , now we substitute this into the second

equation for a:

5

52

3

3

3

60.752.2560.75

60.75 2.2560.752.25

273

ab

bb

b

b

bb

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

=

=

=

Input b = 3, then solve for a 2

2.25a

b= so 2

2.25 2.25 .259

ab

= = = . Since .25a = and b =

3, then ( ) xf x ab= is ( )0.25 3 xy =

4. - 5. A retirement account is opened with an initial deposit of $8,500 and earns 8.12% interest

compounded monthly. What will the account be worth in 20 years? for this we use the formula for compound interest, monthly n = 12

12(20)0.0812( ) 1 8500 1 42888.181212

ntrA t P

n⎛ ⎞ ⎛ ⎞= + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= $42,888.18

6. -

Chapter 6 Review Exercises

600

7. Does the equation 0.6542.294 ty e−= represent continuous growth, continuous decay, or neither? Explain. It is exponential therefore is continuous decay; the growth rate is negative.

8. - Section 6.2

9. Graph the function ( )( ) 3.5 2 .xf x = State the domain and range and give the y-intercept.

The y-intercept is found by inputting x = 0, therefore the y-intercept: ( )0, 3.5 ; For domain we consider all x values we may input . There are no restrictions, therefore the Domain is ( ),−∞ +∞ all real numbers; For the range, we consider the values that y

outputs. The limit on our y-values is the x-axis (it is asymptotic to it) therefore the asymptote is at 0y = and the Range is ( )0,+∞ all real numbers strictly greater than zero;

y-intercept: (0, 3.5).

10. -

11. The graph of ( ) 6.5xf x = is reflected about the y-axis and stretched vertically by a factor of 7. What is the equation of the new function, ( )?g x State its y-intercept, domain, and range.

Chapter 6 Review Exercises

601

For this graph we are asked to reflect it about the y-axis, thus input a negative in front of the x in the exponent thus 6.5 x−= , then stretched vertically means to multiply it by a factor of 7. ( )( ) 7 6.5 ;xg x −

= The y-intercept is found by inputting x = 0,

( )0(0) 7 6.5 7g = = therefore the y-intercept is ( )0, 7 ; For domain we consider all x values

we may input . There are no restrictions, therefore the Domain is ( ),−∞ +∞ all real

numbers; For the range, we consider the values that y outputs. The limit on our y-values is the x-axis (it is asymptotic to it) therefore the asymptote is at 0y = and the Range is

( )0,+∞

( )( ) 7 6.5 ;xg x −= y-intercept: (0, 7); Domain: all real numbers; Range: all real numbers

greater than 0.

12. -

Section 6.3

13. Rewrite ( )17log 4913 x= as an equivalent exponential equation.

Rewriting logbase N exponent= in exponential form is exponentbase N= ; 17 4913x = 14. -

15. Rewrite 25a b

−= as an equivalent logarithmic equation.

Rewriting an exponential expression exponentbase N= in logarithmic form is logbase N exponent= . Since the base a= and the exponent (which always is alone in the

logarithmic equation), 2

5exponent

−= , we rewrite this 2log

5a b = −

16. -

17. Solve for x by converting the logarithmic equation 641log ( )3

x = to exponential form.

Rewriting an exponential expression exponentbase N= in logarithmic form is

logbase N exponent= yields 1

3364 64 4x = = = 18. -

Chapter 6 Review Exercises

602

19. Evaluate ( )log 0.000001 without using a calculator.

Let ( )log 0.000001 ,x= that means 66

1 110 0.000001 101,000,000 10

x −= = = = , that means

6x = − , so ( )log 0.000001 6= −

20. - 21. Evaluate ( )0.8648ln e− without using a calculator

( ) ( )0.8648ln 0.8648ln 0.8648(1) 0.8648e e− = − = − = −

22. - Section 6.4

23. Graph the function ( )( ) log 7 21 4.g x x= + − We must keep

7 21 07 21 21 0 21

7 217 217 7

3

xx

xx

x

+ >

+ − > −

> −

−>

> −

it follows that 3x > − , therefore this graph shifts to the left 3 units, therefore the vertical asymptote which is normally occurs at 0x = , will be at 3x = − . To find the x-intercept we set y = 0:

( )( )( )

4

0 log 7 21 40 4 log 7 21 4 4

4 log 7 2110 7 21

9979 71425

xxx rewrite exponentially

xx

x

= + −

+ = + − +

= +

= +=≈

Therefore the x-intercept would be ( )1425, 0 ;

To find the y-intercept we set x = 0:

( )( )

log 7(0) 21 4.log 21 4.

2.68

yyy

= + −

= −≈ −

Chapter 6 Review Exercises

603

24. -

25. State the domain, vertical asymptote, and end behavior of the function ( )( ) ln 4 20 17.g x x= + −

For the domain, we must keep

4 20 04 20 20 0 20

4 204 204 4

5

xx

xx

x

+ >

+ − > −

> −

−>

> −

it follows that 5x > − , therefore the Domain is ( )5, ;− ∞ This graph shifts to the left 5 units, therefore the vertical asymptote which is normally occurs at 0x = , shifts to the left 5 units. The vertical asymptote will be at 5x = − . Sketching this graph helps describe the end behavior. This is also translated down 17. Domain: 5;x > − Vertical asymptote: 5;x = − End behavior: as x approaches -5 from the right the function values approach negative infinity and as x approaches positive infinity, f(x) increases without bound . Therefore as 5 , ( )x f x+→− →−∞ and as x →∞, f (x)→∞.

Section 6.5

26. - 27. Rewrite ( ) ( ) ( ) ( )8 8 8 8log log 5 log log 13x y+ + + in compact form.

Since addition is going on as individual logarithms, using the product rule in reverse, condensing to a single logarithm will result in multiplication, therefore log8 x( )+ log8 5( )+ log8 y( )+ log8 13( ) = log8 xi5iyi13( ) = log8 65xy( )

28. -

Chapter 6 Review Exercises

604

29. Rewrite ( ) ( ) ( )ln ln lnz x y− − in compact form.

Since subtraction is going on as individual logarithms, using the quotient rule in reverse, condensing to a single logarithm will result in division, therefore grouping the first two

terms we get ( ) ( )( ) ( )ln ln lnz x y− − =

ln

zxy

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

then, since zx

y=

zx

÷y1

=zxi1y

=zxy

our

final answer is

ln

zxy

⎛ ⎞⎜ ⎟

=⎜ ⎟⎜ ⎟⎝ ⎠

ln zxy

⎛ ⎞⎜ ⎟⎝ ⎠

30. -

31. Rewrite 1log12y⎛ ⎞− ⎜ ⎟⎝ ⎠

as a single logarithm.

( ) ( )( )11log log 12 1 1log 1212y y y

−⎛ ⎞− = − = − − =⎜ ⎟⎝ ⎠

( )log 12y

32. -

33. Use properties of logarithms to expand 1ln 2 .1

bb

b⎛ ⎞+⎜ ⎟⎜ ⎟−⎝ ⎠

Rewriting this, and simplifying the numerator we have

ln 2b b+1b−1

⎛

⎝⎜⎜

⎞

⎠⎟⎟. = ln 2bi b+1

b−1⎛

⎝⎜

⎞

⎠⎟

12

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟= ln 2ibi

b+1( )12

b−1( )12

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟= ln

2ibi b+1( )12

b−1( )12

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟ . Now we will apply the

product, quotient rule and power rules

ln2ibi b+1( )

12

b−1( )12

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟= ln 2( )+ ln b( )+ ln b+1( )

12 − ln b−1( )

12 = ln 2( )+ ln b( )+

12

ln b+1( )− 12

ln b−1( ) =

( ) ( ) ( ) ( )ln 1 ln 1ln 2 ln

2b b

b+ − −

+ +

34. -

35. Condense the expression 77 7

log3log 6log

3u

v w+ − to a single logarithm.

Since there are coefficients on the front of the logarithms we apply the power rule first

(the last term has 13

as a coefficient) and place them as exponents

Chapter 6 Review Exercises

605

( ) ( )1

3 67 37 7 7 7 7

log3log 6log log log log3

uv w v w u⎛ ⎞+ − = + + ⎜ ⎟

⎝ ⎠. Since addition is going on as

individual logarithms and subtraction of the last term, using the product & quotient rule in reverse, condensing to a single logarithm will result in multiplication then division, therefore

log7 v3( )+ log7 w6( )+ log7 u13

⎛

⎝⎜

⎞

⎠⎟= log7

v4 iw6

u13

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟=

3 6

7 3log v w

u

⎛ ⎞⎜ ⎟⎝ ⎠

36. -

37. Rewrite 12 175 125x− = as a logarithm. Then apply the change of base formula to solve for x using the common log. Round to the nearest thousandth.

12 17

5

5 125log 125 12 17log125 12 17log5

log125 1712log5

12 12log125 17log5

12

x

x

x

x

x

− == −

= −

+=

+=

( )( )

log 12517

log 5 512 3

x+

= =

Section 6.6

38. -

39. Solve 33

125 51

625

x− −=

⎛ ⎞⎜ ⎟⎝ ⎠

by rewriting each side with a common base.

Find the same common base between 125, 625 and 5 which would be 5. We want to rewrite each side with the base of 5, using all the correct exponent rules. Once we have a single base expression on each side and the base is the same, we set the exponents equal and solve for x

Chapter 6 Review Exercises

606

( )( )

33

33

34

33

4 12

3 (4 12) 3

3 4 12 3

125 51

6255 5 Rewrite each side as a power with base 5.

5

5 5 subtract exponents on left side .5

5 5 5 54 9 3 Apply the one-to-one property of expo

x

x

x

x

x

x

− −

− −−

+

− +

− −

=⎛ ⎞⎜ ⎟⎝ ⎠

=

=

==

− − = nents.4 12 Subtract 12 from both sides.

3 Divide by -4.xx

− == −

3x = − 40. - 41. Use logarithms to find the exact solution for 6 23 1 60.ne − + = − If there is no solution, write

no solution. We want the 6 2ne − alone on one side before we begin.

6 2

6 2

6 2

6 2

3 1 60. Subtract 1 from both sides 3 61 Divide both sides by 3.

61 3

61ln ln Take ln of both sides.3

n

n

n

n

e

e

e

e

−

−

−

−

+ = −

= −

−=

−⎛ ⎞= ⎜ ⎟⎝ ⎠

We can stop here because 61ln3−⎛ ⎞

⎜ ⎟⎝ ⎠

is not defined, therefore there is no solution

42. -

43. Find the exact solution for 5 22 9 56.xe − − = − If there is no solution, write no solution. We want the 5 2xe − alone on one side before we begin.

Chapter 6 Review Exercises

607

5 2

5 2

5 2

5 2

2 9 56 Add 9 to both sides 2 47 Divide both sides by 2.

47 2

47ln ln Take ln of both sides.2

x

x

x

x

e

e

e

e

−

−

−

−

− = −

= −

−=

−⎛ ⎞= ⎜ ⎟⎝ ⎠

We can stop here because 47ln 2−⎛ ⎞⎜ ⎟⎝ ⎠

is not defined, therefore there is no solution

44. - 45. Find the exact solution for 2 110 0.x xe e− − = If there is no solution, write no solution.

Since this is a quadratic form of an exponential equation , we plan to use factoring to solve the problem. We always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero.

2 110 0. Get one side of the equation equal to zero.( 10)( 11) 0 Factor by the FOIL method.

10 0 or 11 0 If a produ

x x

x x

x x

e e

e e

e e

− − =

+ − =

+ = − =

( )

ct is zero, then one factor must be zero.10 or e 11 Isolate the exponentials.

11 Reject the equation in which the power equals a negative number.ln 11

x x

x

e

e

x

= − =

=

= Solve the equation in which the power equals a positive number.

( )ln 11x =

46. -

47. Use the definition of a logarithm to find the exact solution for ( )9 6ln 3 33.a+ + =

We want the ( )ln 3a + alone on one side before we begin.

( )( )( )

4

4

9 6ln 3 33 Subtract 9 from both sides6ln 3 24 Divide both sides by 6

ln 3 43 Rewrite exponentially

3 Subtract 3 from both sides

aaa

e ae a

+ + =

+ =

+ =

= +− =

Substituting 4 3e a− = into the original logarithmic functions we see the argument of the logarithm functions is positive therefore this is a solution.

4 3a e= − 48. -

Chapter 6 Review Exercises

608

49. Use the one-to-one property of logarithms to find an exact solution for ( ) ( ) ( )2ln 5 ln 5 5 ln 56 .x+ − = If there is no solution, write no solution.

We have to rewrite each side as a single logarithm first ( ) ( ) ( )

( )( ) ( )

2

2

2

2

2

2

ln 5 ln 5 5 ln 56

ln 5 5 5 ln 56

25 25 56 Use the one-to-one property of the logarithm25 =81 Add 25 to both sides25 81 Divide by 25

81 Take the square root of both sides25

95

x

x

xxx

x

x

+ − =

− =

− =

=

=

= ±

g

Substituting both 95

x = ± into the original logarithmic functions we see the argument of

the logarithm functions is positive therefore each one is a solution. 95

x = ±

50. - 51. The population of a city is modeled by the equation 0.25( ) 256,114 tP t e= where t is

measured in years. If the city continues to grow at this rate, how many years will it take for the population to reach one million? We are being asked to solve for t when ( ) 1,000,000P t =

( ) ( )

0.25

0.25 0.25

0.25

0.25

0.25

( ) 256,1141,000,000 256,114 alone on one side1,000,000256,114

3.904511272 take ln of both sides to solve for xln 3.904511272 ln recall power rule of logs on righ

t

t t

t

t

t

P t ee get e

e

ee

==

=

=

=

( ) ( )( )( )

( )

t side and ln e = 1ln 3.904511272 0.25 lnln 3.904511272 0.25 Divide by 0.25ln 3.904511272

0.25ln 3.904511272 0.25 5.448530484

t et

xt

x

=

=

=

= ÷ =

It will take about 5.45 years 52. -

53. Find the inverse function 1f − for the logarithmic function f x( ) = 0.25⋅ log2 x3 +1( ).

Chapter 6 Review Exercises

609

( )320.25 log 1y x= ⋅ + We solve for x so we want to get the logarithm is alone on one

side, so we can rewrite this in exponential form

( )( )( )

32

32

32

4 3

4 3

0.25 log 1 get log alone on right side, divide by 0.251log 1 1 .25 4

0.25 .254 log 1 rewrite exponentially

2 1 subtract 12 1 take the cube root of b

y

y

y xy

x

y x

xx

= ⋅ +

= + = ÷ =

= +

= +− =

3 4

oth sides2 1y x− =

Where there is the y we place x, so the inverse is ( ) 31 42 1xf x− = −

Section 6.7 For the following exercises, use this scenario: A doctor prescribes 300 milligrams of a therapeutic drug that decays by about 17% each hour.

54. - 55. Write an exponential model representing the amount of the drug remaining in the

patient’s system after t hours. Then use the formula to find the amount of the drug that would remain in the patient’s system after 24 hours. Round to the nearest hundredth of a gram. Referring to the previous problem we found 300(.83)x xy ab= = , now we plug in t = 24

hours 24(24) 300(.83) 3.427642126f = =

( )( ) 300 0.83 ; (24) 3.43tf t f g= ≈

For the following exercises, use this scenario: An oven brick with a temperature of 350° Fahrenheit was taken off the stove to cool in a 71 F° room. After fifteen minutes, the temperature of the brick was 175 F.°

56. -

57. How many minutes will it take the brick to cool to 85 F?° To solve this we use our model and T(t)=85, and solve for t.

Chapter 6 Review Exercises

610

( 0.065788 )

( 0.065788 )

( 0.065788 )

0.065788

( ) 279 71,85 ( ) 279 71, subtract 71, divide by 279

14 279279 279

7 Take the ln of both sides, power rule on right side and ln e = 118

14ln279

t

t

t

t

T t eT t e

e

e

−

−

−

−

= += = +

=

=

⎛ ⎞ =⎜ ⎟⎝ ⎠

0.065788 divide by 0.065788

14ln279

0.06578845.48172581

t

t

t

− −

⎛ ⎞⎜ ⎟⎝ ⎠ =

−=

The brick will cool to 85 degrees in about 45 minutes

For the following exercises, use this scenario: The equation ( ) 0.625

12001 199 tN t

e−=

+ models the

number of people in a school who have heard a rumor after t days. 58. - 59. To the nearest tenth, how many days will it be before the rumor spreads to half the

carrying capacity? Half the carrying capacity is half of 1200 = 600, so we will solve for t

( )0.625

0.625

0.625

0.625

0.625

12006001 199

600 1 199 1200 divide by 600

1 199 2199 1

1 take ln of both sides , use exponents rules199

1ln 0.625 divide by 0.6251991ln

1990

t

t

t

t

t

ee

ee

e

t

−

−

−

−

−

=+

+ =

+ ==

=

⎛ ⎞ = − −⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟⎝ ⎠− .625

8.46928772

t

t

=

=

In about 8.5 days.

60. –

Chapter 6 Review Exercises

611

61. x 1 2 3 4 5 6 7 8 9 10

f(x) 3.05 4.42 6.4 9.28 13.46 19.52 28.3 41.04 59.5 86.28

Using a graphing calculator, you can plot individual points using STAT Plot. Press 2nd STAT PLOT, which is above Y=. Highlight 1:Plot1…On, Enter, Choose what type of mark you’d like , enter. To enter coordinates hit the STAT button, 1:Edit, enter. It brings L1 and L2 and L3. Enter the x values down column L1 and then the corresponding y-values in L2 column. Once they are all in exit that screen and hit GRAPH. It should plot the points for you. You have to go to the y= menu and make sure you highlight Plot1 above the Y1 symbol. Also Use ZOOM [9] to adjust axes to fit the data.

Because of the concave up curve it is exponential.

62. -

63. Find a formula for an exponential equation that goes through the points ( )2, 100− and

( )0, 4 . Then express the formula as an equivalent equation with base e. We are given the point (0,4) and since

0

( )44

xf x ab

aba

=

=

=

We solve for b by plugging in the given point (-2,100)

Chapter 6 Review Exercises

612

2

2

2

2

2

100 4( )25 divide by 4

125=

25 11251 .2 choose the positive number5

b

b

bb

b

b

−

−

=

=

=

=

= ± =

( )4 0.2 ;xy = to change it to base e ( ) ( )ln.24 0.2 4x xxy ab e so= = = -1.6094384 xy e=

Section 6.8

64. -

65. The population of a culture of bacteria is modeled by the logistic equation

0.62

14,250( ) ,1 29 tP t

e−=

+ where t is in days. To the nearest tenth, how many days will it take the

culture to reach 75% of its carrying capacity? 75% of the carrying capacity would be .75(14,2500) 10,687.5= . We will solve the equation for t when it’s equal to 10,687.5:

( )0.62

0.62

0.62

0.62

0.62

14,25010,687.5 ,1 29

10,687.5 1 29 14,250 divide by 10,687.541 293129 divide by 2931 take ln of both sides , use exponents rules

871ln 0.62 divid

87

t

t

t

t

t

ee

e

e

e

t

−

−

−

−

−

=+

+ =

+ =

=

=

⎛ ⎞ = −⎜ ⎟⎝ ⎠

e by 0.62

1ln87

0.627.203077611

t

t

−

⎛ ⎞⎜ ⎟⎝ ⎠ =−

=

So it will be in about 7.2 days .

Chapter 6 Review Exercises

613

For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best described by an exponential, logarithmic, or logistic model. Then use the appropriate regression feature to find an equation that models the data. When necessary, round values to five decimal places.

66. - 67.

x 0.15 0.25 0.5 0.75 1 1.5 2 2.25 2.75 3 3.5

f(x) 36.21 28.88 24.39 18.28 16.5 12.99 9.91 8.57 7.23 5.99 4.81

Using a graphing calculator, you can plot individual points using STAT Plot. Press 2nd STAT PLOT, which is above Y=. Highlight 1:Plot1…On, Enter, Choose what type of mark you’d like, enter. To enter coordinates hit the STAT button, 1:Edit, enter. It brings L1 and L2 and L3. Enter the x values down column L1 and then the corresponding y-values in L2 column. Once they are all in exit that screen and hit GRAPH. It should plot the points for you. You have to go to the y= menu and make sure you highlight Plot1 above the Y1 symbol. Also Use ZOOM [9] to adjust axes to fit the data.

Because it appear the data is asymptotic to the y-axis as it gets close to x=0, we would choose the logarithmic curve. To find the logarithmic equation that models the data, select “LnReg” from the STAT then CALC menu. Use the values returned for a and b to record the model, ( )lny a b x= + Your screen will display

ln16.68718406

9.718602513

y a b xab

= +

=

= −

Therefore the logarithmic function that best fits the data in the table is

16.68718 9.71860ln( )y x= −

68. -

Chapter6PracticeTest

614

Chapter 6 Practice Test 1. The population of a pod of bottlenose dolphins is modeled by the function

( ) 8(1.17) ,tA t = where t is given in years. To the nearest whole number, what will the pod population be after 3 years? Since ( ) 8(1.17) ,tA t = we want to find 3(3) 8(1.17) 12.812904 13A = = ≈ .

About 13 dolphins. 2. - 3. Drew wants to save $2,500 to go to the next World Cup. To the nearest dollar, how

much will he need to invest in an account now with 6.25% APR, compounding daily, in order to reach his goal in 4 years? For this we use the Present Value formula derived from the compound interest formula, compounded daily n= 365

365(4)0.0625( ) 1 2500 1 1947.045627

365

ntrP A t

n

− −⎛ ⎞ ⎛ ⎞= ⋅ + = ⋅ + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

. He will have to invest

approximately $1947 . 4. - 5. Graph the function ( )( ) 5 0.5 xf x −

= and its reflection across the y-axis on the same axes, and give the y-intercept. To graph this reflected about the y-axis puts a negative in front of the – x in the exponent. This would make the new function ( ) ( )( )( ) 5 0.5 5 0.5x xf x − −

= = . The

y-intercept is still (0, 5) .

Chapter6PracticeTest

615

6. - 7. Rewrite ( )8.5log 614.125 a= as an equivalent exponential equation.

Rewriting logbase N exponent= in exponential form is exponentbase N= ;

8.5 614.125a = 8. -

9. Solve for x by converting the logarithmic equation 1

7

log ( ) 2x = to exponential form.

Rewriting logbase N exponent= in exponential form is exponentbase N= yields 21 1

7 49x ⎛ ⎞= =⎜ ⎟

⎝ ⎠

10. - 11. Evaluate ( )ln 0.716 using a calculator. Round to the nearest thousandth.

( )ln 0.716 0.334≈ −

12. - 13. State the domain, vertical asymptote, and end behavior of the function

( )5( ) log 39 13 7.f x x= − +

For the Domain we must keep

39 13 039 13 39 0 39

13 3913 3913 13

3

xx

xx

x

− >

− − > −

− > −

− −<

− −<

it follows that 3x < , therefore the Domain is ( ) ,3 ;−∞ This graph shifts to the right 3 units, therefore the vertical asymptote which is normally occurs at 0x = , shifts to the right 3 units. The vertical asymptote will be 3;x = Sketching this graph helps describe the end behavior. As x approaches 3 from the left the end behavior: as

3 , ( )x f x−→ →−∞ and as , ( )x f x→−∞ →∞

Chapter6PracticeTest

616

14. - 15. Rewrite ( ) ( )log 96 log 8t t− in compact form.

Since subtraction is going on as individual logarithms, using the quotient rule in reverse, condensing to a single logarithm will result in division

( ) ( ) 96log 96 log 8 log8t t t

⎛ ⎞− = ⎜ ⎟⎝ ⎠

= ( )log 12t

16. - 17. Use properties of logarithm to expand ( )3 2 3ln 4 .y z x⋅ −

apply the product, and power rules

ln y3 ⋅ z2 ⋅ x − 43( ) = ln y3iz2 i x − 4( )13

⎛

⎝⎜

⎞

⎠⎟= ln y3( )+ ln z2( )+ ln x − 4( )

13 = 3ln y( )+ 2ln z( )+

13

ln x − 4( ) =

( ) ( ) ( )ln 43ln 2ln

3x

y z−

+ +

18. - 19. Rewrite 3 516 1000x− = as a logarithm. Then apply the change of base formula to solve

for x using the natural log. Round to the nearest thousandth. 3 5

16

16 1000log 1000 3 5

ln1000 3 5ln16

ln1000 5 3ln163 3

ln1000 5ln16

3

x

x

x

x

x

− == −

= −

+=

+=

( )( )

ln 10005

ln 162.497

3x

+

= ≈

20. - 21. Use logarithms to find the exact solution for 10 89 5 41ae −− − = − . If there is no solution,

write no solution.

Chapter6PracticeTest

617

We want the 10 8ae − alone on one side before we begin.

( )( ) ( )

10 8

10 8

10 8

10 8

9 5 41 Add 5 to both sides9 36 4 Divide both sides by -9.

ln ln 4 Take ln of both sides.

10 8 ln ln 4 Use law

a

a

a

a

e

e

e

e

a e

−

−

−

−

− − = −

− = −

=

=

− =

( ) ( )( )( )

s of logs. Remember ln e = 1

10 8 (1) ln 4

10 8 ln 4 Add 8 and Divide both sides by 10

ln 4 8

10

a

a

a

− =

− =

+=

22. - 23. Find the exact solution for 4 15 4 64.xe− −− − = If there is no solution, write no solution.

We want the 4 1xe− − alone on one side before we begin.

( ) ( )

4 1

4 1

4 1

4 1

5 4 64 Add 4 to both sides5 60 Divide both sides by -5.

60 5

ln ln 12 Take ln of both sides.

x

x

x

x

e

e

e

e

− −

− −

− −

− −

− − =

− =

=−

= −

We can stop here because ( )ln 12− is not defined, therefore there is No solution

24. - 25. Find the exact solution for 2 72 0.x xe e− − = If there is no solution, write no solution.

Since this is a quadratic form of an exponential equation, we plan to use factoring to solve the problem. We always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero.

Chapter6PracticeTest

618

2 72 0 Get one side of the equation equal to zero.( 8)( 9) 0 Factor by the FOIL method.

8 0 or 9 0 If a product is

x x

x x

x x

e e

e e

e e

− − =

+ − =

+ = − =

( )

zero, then one factor must be zero.8 or e 9 Isolate the exponentials.

9 Reject the equation in which the power equals a negative number.ln 9

x x

x

e

e

x

= − =

=

= Solve the equation in which the power equals a positive number.

26. - 27. Use the one-to-one property of logarithms to find an exact solution for

( ) ( ) ( )2log 4 10 log 3 log 51x − + = If there is no solution, write no solution.

We have to rewrite each side as a single logarithm first

( ) ( ) ( )

( )( ) ( )

2

2

2

2

2

2

log 4 10 log 3 log 51

log 4 10 3 ln 51

12 30 51 Use the one-to-one property of the logarithm12 81 Divide by 12

811227 Take the square root of both sides4

27 3 32 2

x

x

xx

x

x

x

− + =

− =

− ==

=

=

= ± = ±

g

Substituting both 3 32

x = ± into the original logarithmic functions we see the

argument of the logarithm functions is positive therefore each one is a solution.

3 32

x = ±

28. -

Chapter6PracticeTest

619

29. A radiation safety officer is working with 112 grams of a radioactive substance. After 17 days, the sample has decayed to 80 grams. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest day, what is the half-life of this substance? Since the initial amount 0 112A g= and after 17 days, it decays to 80 grams we need to find k, so plugging in the given information we see:

0 0

(17 )

17

80 112 divide both sides by 112 to get e0.7142857143 take the ln of both sides and use power rule on rigft side

ln 0.7142857143 17 ln ln e = 1ln 0.7142857143 17 d

kt

k days kt

k

A A e

e alone

ek ek

=

=

=

=

=

.019792

ivide both sides by 17ln 0.7142857143

17.0197924845 .01979

( ) 112 t

k

k

Therefore f t e−

=

= − ≈ −

=

To find the half-life, ln 2 ln 2 35.0215835 35 days.01979

tk

− −= = = ≈

−

30. - 31. A bottle of soda with a temperature of 71° Fahrenheit was taken off a shelf and placed in

a refrigerator with an internal temperature of 35 F.° After ten minutes, the internal temperature of the soda was 63 F.° Use Newton’s Law of Cooling to write a formula that models this situation. To the nearest degree, what will the temperature of the soda be after one hour?

The model formula we use is ( ) ktsT t Ae T= + and we know 35sT = ° and

71 35 36A = − = , therefore to find k, we plug in the fact that when 10min, (10) 63t T= =

Chapter6PracticeTest

620

(10)

(10)

10

( )

63 36 35 Subtract 35 and then divide by 3628 3636 367 Take the ln of both sides, power rule on right side and ln e = 19

7ln 10 9

kts

k

k

k

T t Ae T

e

e

e

k

= +

= +

=

=

⎛ ⎞ =⎜ ⎟⎝ ⎠

divide by 10

7ln9

100.0251314428

k

k

⎛ ⎞⎜ ⎟⎝ ⎠ =

≈ −

Therefore our model would be: 0.025131( ) 36 35,tT t e−= + Using our model to find the temperature after 1 hours we will plug in t= 60 minutes

0.025131(60)(60) 36 35 42.96979633 43T e F−= + = ≈ ° 32. - 33. Enter the data from the table below into a graphing calculator and graph the resulting

scatter plot. Determine whether the data from the table would likely represent a function that is linear, exponential, or logarithmic.

x 1 2 3 4 5 6 7 8 9 10

f(x) 3 8.55 11.7

9 14.0

9 15.8

8 17.3

3 18.5

7 19.6

4 20.5

8 21.4

2

So Using a graphing calculator, you can plot individual points using STAT Plot. Press 2nd STAT PLOT, which is above Y=. Highlight 1:Plot1…On, Enter, Choose what type of mark you’d like , enter. To enter coordinates hit the STAT button, 1:Edit, enter. It brings L1 and L2 and L3. Enter the x values down column L1 and then the corresponding y-values in L2 column. Once they are all in exit that screen and hit GRAPH. It should plot the points for you. You have to go to the y= menu and make sure you highlight Plot1 above the Y1 symbol. Also Use ZOOM [9] to adjust axes to fit the data. This is a logarithmic curve.

Chapter6PracticeTest

621

34. -

35. x 1 2 3 4 5 6 7 8 9 10

f(x) 20 21.6 29.2 36.4 46.6 55.7 72.6 87.1 107.2 138.1

So Using a graphing calculator, you can plot individual points using STAT Plot. Press 2nd STAT PLOT, which is above Y=. Highlight 1:Plot1…On, Enter, Choose what type of mark you’d like , enter. To enter coordinates hit the STAT button, 1:Edit, enter. It brings L1 and L2 and L3. Enter the x values down column L1 and then the corresponding y-values in L2 column. Once they are all in exit that screen and hit GRAPH. It should plot the points for you. You have to go to the

Chapter6PracticeTest

622

y= menu and make sure you highlight Plot1 above the Y1 symbol. Also Use ZOOM [9] to adjust axes to fit the data. This is an exponential curve.

To Find the equation that models the data. Select “ExpReg” from the STAT then CALC menu. Use the values returned for a and b to record the model, .xy ab= Your screen will display

* ^15.100622021.246214498

y a b xab

=

=

=

Therefore the exponential function that best fits the data in the table is ( )15.10062 1.24621 xy =

36. -

Chapter6PracticeTest

623

37. x 0 0.5 1 1.5 2 3 4 5 6 7 8

f(x) 2.2 2.9 3.9 4.8 6.4 9.3 12.3 15 16.2 17.3 17.9

This has the shape of a logistic curve. Therefore we will use: To find the LOGISTIC equation that models the data. Select “B:Logistic” from the STAT then CALC menu. Use the values returned for a , b and c to record the model,

1 bx

cy

ae−=

+

Your screen will display / (1 ^ ( ))

7.546440174.683750473618.41658808

y c ae bxabc

= + −

=

=

=

Therefore the LOGISTIC function that best fits the data in the table is

0.68375

18.416591 7.54644 xy

e−=

+

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